
Chapter 12 – Simple Harmonic Motion
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Question 16:
In figure (12−E3) k = 100 N m^{−1}, M = 1 kg and F = 10 N. (a) Find the compression of the spring in the equilibrium position. (b) A sharp blow by some external agent imparts a speed of 2 m s^{−1} to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant. (c) Find the time period of the resulting simple harmonic motion. (d) Find the amplitude. (e) Write the potential energy of the spring when the block is at the left extreme. (f) Write the potential energy of the spring when the block is at the right extreme.
The answer of (b), (e) and (f) are different. Explain why this does not violate the principle of conservation of energy.
Figure
Answer:
It is given that:
Spring constant, k = 100 N/m,
Mass of the block, M = 1 kg
Force, F = 10 N
(a) In the equilibrium position,
F=kx where x is the compression of the spring, and
k is the spring constant.
∴ x=Fk=10100 =0.1 m=10 cm(b) The blow imparts a speed of 2 ms^{1} to the block, towards left.
Potential energy of spring, U =
12kx2 Kinetic energy,
K=
12Mv2(c) Time period
Tis given by,
T = 2πMk
=2π1100=π5 s(d) Let A be the amplitude.
Amplitude is the distance between the mean and the extreme position.
At the extreme position, compression of the spring will be (A + x).
As the total energy in S.H.M. remains constant, we can write:
12kA+x2=12kx2+12Mv2+Fx =2.5+10x ∴ 50(A + 0.1)^{2} = 2.5 + 10x
⇒50A^{2} + 0.5 + 10A = 2.5 + 10A
⇒50A^{2} = 2
⇒A2=250=4100⇒A=210 m=0.2 m=20 cm(e) Potential Energy at the left extreme will be,
P.E.=12kA+x2 =12×100×0.1+0.22 =50×0.09=4.5 J(f) Potential Energy at the right extreme is calculated as:
Distance between the two extremes = 2A
P.E.=12kA+x2F2A
= 4.5 − 10 (0.4) = 0.5 J
As the work is done by the external force of 10 N, different values of options (b), (e) and (f) do not violate the law of conservation of energy.
Question 17:
Find the time period of the oscillation of mass m in figures 12−E4 a, b, c. What is the equivalent spring constant of the pair of springs in each case?
Figure
Answer:
(a) Spring constant of a parallel combination of springs is given as,
K = k_{1} + k_{2} (parallel)
Using the relation of time period for S.H.M. for the given springmass system, we have:
T=2πmK=2πmk1+k2
(b) Let x be the displacement of the block of mass m, towards left.
Resultant force is calculated as,
F = F_{1} + F_{2} = (k_{1} + k_{2})x
Acceleration
ais given by,
a=Fm=k1+k2mxTime period
Tis given by,
T=2πdisplacementaccelerationOn substituting the values of displacement and acceleration, we get:T =2πxxk1+k2m =2πmk1+k2
Required spring constant, K = k_{1} + k_{2}
(c) Let K be the equivalent spring constant of the series combination.
1K=1k1+1k2=k2+k1k1k2⇒K=k1k2k1+k2Time period is given by,T=2πmKOn substituting the respective values, we get:T=2πmk1+k2k1k2
Question 18:
The spring shown in figure (12−E5) is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.
Figure
Answer:
(a) We know
f = kx
⇒x=FkAcceleration=FmUsing the relation of time period of S.H.M.,
.
Time period, T=2πDisplacementAcceleration =2πFkFm=2πmkAmplitude = Maximum displacement
=FkWhen the block passes through the equilibrium position, the energy contained by the spring is given by,
E=12kx2=12kFk2=12F2k(b) At the mean position, potential energy is zero.
Kinetic energy is given by,
12kx2=12F2k
Question 19:
A particle of mass m is attatched to three springs A, B and C of equal force constants k as shown in figure (12−E6). If the particle is pushed slightly against the spring C and released, find the time period of oscillation.
Figure
Answer:
(a) Let us push the particle lightly against the spring C through displacement x.
As a result of this movement, the resultant force on the particle is kxâ€‹.
The force on the particle due to springs A and B is
kx2.
Total Resultant force
=kx+kx22+kx22 = kx + kx = 2kx
Acceleration is given by
=2kxm
Time period=2πDisplacementAcceleration =2πx2kx/m =2πm2k
Question 20:
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
Answer:
As the particle is pushed against the spring C by the distance x, it experiences a force of magnitude kx.
If the angle between each pair of the springs is 120Ëš then the net force applied by the springs A and B is given as,
kx22+kx22+2kx2kx2 cos 120° =kx2Total resultant force
Facting on mass m will be,
F=kx+kx2=3kx2 ∴a=Fm=3kx2m⇒ax=3k2m=ω2⇒ω=3k2m∴Time period, T=2πω=2π2m3k
Question 21:
The springs shown in the figure (12−E7) are all unstretched in the beginning when a man starts pulling the block/ The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.
Figure
Answer:
As the block of mass M is pulled, a net resultant force is exerted by the three springs opposing the motion of the block.
Now, springs k_{2} and k_{3} are in connected as a series combination.
Let k_{4} be the equivalent spring constant.
∴1k4=1k2+1k3=k2+k3k2k3k4=k2k3k2+k3k_{4} and k_{1} form a parallel combination of springs. Hence, equivalent spring constant k = k_{1} + k_{4}.
=k2k3k2+k3+k1=k2k3+k1k2+k1k3k2+k3∴Time peiod, T=2πMk =2πM k2+k3k2k3+k1k2+k1k3(b) Frequency
vis given by,
v=1T
=12πk2k3+k1k2+k1k3Mk2+k3(c) Amplitude ( x ) is given by,
x=Fk=Fk2+k3k1k2+k2k3+k1k3
Question 22:
Find the elastic potential energy stored in each spring shown in figure (12−E8), when the block is in equilibrium. Also find the time period of vertical oscillation of the block.
Figure
Answer:
All three spring attached to the mass M are in series.
k_{1}, k_{2}, k_{3} are the spring constants.
Let k be the resultant spring constant.
1k=1k1+1k2+1k3⇒k=k1k2k3k1k2+k2k3+k3k1Time period T is given by,T=2πMk =2Mk1k2+k2k3+k3k1k1k2k3 =2M1k1+1k2+1k3As force is equal to the weight of the body,
F = weight = Mg
Let x_{1}_{, }x_{2}, and x_{3} be the displacements of the springs having spring constants k_{1}, k_{2} and k_{3} respectively.
â€‹For spring k_{1,}
x1=Mgk1Similarly, x2=Mgk2and x3=Mgk3∴PE1=12k1x12 =12k1Mgk12 =12k1M2g2k12 =12M2g2k1=M2g22k1Similarly, PE2=M2g22k2and PE3=M2g22k3
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Question 23:
The string, the spring and the pulley shown in figure (12−E9) are light. Find the time period of the mass m.
Figure
Answer:
Let l be the extension in the spring when mass m is hung.
Let T_{1} be the tension in the string; its value is given by,
T_{1}_{ =} kl = mg
Let x be the extension in the string on applying a force F.
Then, the new value of tension T_{2} is given by,
T_{2} = k(x + l)
Driving force is the difference between tensions T_{1} and T_{2}_{.}
∴ Driving force = T_{2} − T_{1} = k(x + l) − kl
= kx
Acceleration, a=kxmTime period T is given by,T=2πdisplacementAcceleration =2πxkx/m=2πmk
Question 24:
Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.
Answer:
Let us try to solve the problem using energy method.
If δ is the displacement from the mean position then, the initial extension of the spring from the mean position is given by,
δ = mgkLet x be any position below the equilibrium during oscillation.
Let v be the velocity of mass m and ω be the angular velocity of the pulley.
If r is the radius of the pulley then
v = rω
As total energy remains constant for simple harmonic motion, we can write:
12Mv2+12Iω2+12k(x+δ)2δ2Mgx=Constant⇒12Mv2+12Iω2+12kx2+kxdMgx=Constant⇒12Mv2+12Iv2r2+12kx2=Constant ∵δ=MgkBy taking derivatives with respect to t, on both sides, we have:
Mv.dvdt+Ir2v.dvdt+kxdxdt=0Mva+Ir2va+kxv=0 ∵v=dxdt and a=dvdtaM+Ir2=kx ⇒ax=kM+Ir2=ω2T=2πω⇒T=2πM+Ir2k
Question 25:
Consider the situation shown in figure (12−E10). Show that if the blocks are displaced slightly in opposite direction and released, they will execute simple harmonic motion. Calculate the time period.
Figure
Answer:
The centre of mass of the system should not change during simple harmonic motion.
Therefore, if the block m on the left hand side moves towards right by distance x, the block on the right hand side should also move towards left by distance x. The total compression of the spring is 2x.
If v is the velocity of the block. Then
Using energy method, we can write:
12k2x2+12mv2+12mv2=C⇒ mv^{2} + 2kx^{2} = C
By taking the derivative of both sides with respect to t, we get:
2mvdvdt+2k×2xdxdt=0Putting v=dxdt;and a=dvdtin above expression, we get ma+2kx=0 ⇒ax=2km=ω2⇒ω=2km⇒Time period, T=2πm2k
Question 26:
A rectangle plate of sides a and b is suspended from a ceiling by two parallel string of length L each (figure 12−E11). The separation between the string is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period.
Figure
Answer:
Let m is the mass of rectangular plate and x is the displacement of the rectangular plate.
During the oscillation, the centre of mass does not change.
Driving force
Fis given as,
F = mgsin θ
Comparing the above equation with F = ma, we get:
a = Fm = gsinθFor small values of θ, sinθ can be taken as equal to θ.
Thus, the above equation reduces to:
a=gθ=gxL Where g and L are constant.It can be seen from the above equation that, a α x.
Hence, the motion is simple harmonic.
Time period of simple harmonic motion
Tis given by,
T=2πdisplacementAcceleration =2πxgx/L=2πLg
Question 27:
A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N m^{−1}. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.
Figure
Answer:
It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Total mass of the system, M = 3 + 1 = 4 kg (when both the blocks move together)
Spring constant, k = 100 N/m
â€‹Time period of SHM
T is given by,
T=2πMkOn substituting the values of M and k in the bove equation, we have: T=2π4100=2π5 sFrequency of the motion is given by, 1T=52π HzLet v be the velocity of the 1 kg block, at mean position.
As kinetic energy is equal to the potential energy, we can write:12mv2=12kx2where x = amplitude = 0.1 m
Substituting the value of x in above equation and solving for v, we get:12×1×v2=12×1000.12 v=1 ms1 …1When the 3 kg block is gently placed on the 1 kg block, the 4 kg mass and the spring become one system. As a springmass system experiences external force, momentum should be conserved.
Let V be the velocity of 4 kg block.
Now,
Initial momentum = Final momentum
∴ 1 × v = 4 × V
⇒V=14 m/s As v=1 ms1, from equation (1)Thus, at the mean position, two blocks have a velocity of
14ms1.
Mean value of kinetic energy is given as,KE at mean position=12MV2 =12×4×142=12×14=18At the extreme position, the springmass system has only potential energy.
PE=12kδ2=12×14where δ is the new amplitude.
∴14=100 δ2 =δ=1400 =0.05 m = 5 cm
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Question 46:
A simple pendulum of length l is suspended from the ceiling of a car moving with a speed v on a circular horizontal road of radius r. (a) Find the tension in the string when it is at rest with respect to the car. (b) Find the time period of small oscillation.
Answer:
It is given that a car is moving with speed v on a circular horizontal road of radius r.
(a) Let T be the tension in the string.
According to the free body diagram, the value of T is given as,
T=mg2+mv2r2
=mg2+v4r2=ma, where acceleration, a
=g2+v4r2The time period
Tis given by,
T=2πlgOn substituting the respective values, we have:T=2πlg2+v4r21/2
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Question 1:
A person goes to bed at sharp 10.00 pm every day. Is it an example of periodic motion? If yes, what is the time period? If no, why?
Answer:
No. As motion is a change in position of an object with respect to time or a reference point, it is not an example of periodic motion.
Question 2:
A particle executing simple harmonic motion comes to rest at the extreme positions. Is the resultant force on the particle zero at these positions according to Newton’s first law?
Answer:
No. The resultant force on the particle is maximum at the extreme positions.
Question 3:
Can simple harmonic motion take place in a noninertial frame? If yes, should the ratio of the force applied with the displacement be constant?
Answer:
Yes. Simple harmonic motion can take place in a noninertial frame. However, the ratio of the force applied to the displacement cannot be constant because a noninertial frame has some acceleration with respect to the inertial frame. Therefore, a fictitious force should be added to explain the motion.
Question 4:
A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is its displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?
Answer:
No, we cannot say anything from the given information. To determine the displacement of the particle using its velocity at any instant, its mean position has to be known.
Question 5:
A small creature moves with constant speed in a vertical circle on a bright day. Does its shadow formed by the sun on a horizontal plane move in a sample harmonic motion?
Answer:
Yes, its shadow on a horizontal plane moves in simple harmonic motion. The projection of a uniform circular motion executes simple harmonic motion along its diameter (which is the shadow on the horizontal plane), with the mean position lying at the centre of the circle.
Question 6:
A particle executes simple harmonic motion Let P be a point near the mean position and Q be a point near an extreme. The speed of the particle at P is larger than the speed at Q. Still the particle crosses P and Q equal number of times in a given time interval. Does it make you unhappy?
Answer:
No. It does not make me unhappy because the number of times a particle crosses the mean and extreme positions does not depend on the speed of the particle.
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Question 7:
In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.
Answer:
The mean position of a particle executing simple harmonic motion is fixed, whereas its extreme position keeps on changing. Therefore, when we use stop watch to measure the time between consecutive passage, we are certain about the mean position.
Question 8:
It is proposed to move a particle in simple harmonic motion on a rough horizontal surface by applying an external force along the line of motion. Sketch the graph of the applied force against the position of the particle. Note that the applied force has two values for a given position depending on whether the particle is moving in positive or negative direction.
Answer:
Figure (a) shows the graph of the applied force against the position of the particle.
(a)
(b)
(c)
Question 9:
Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?
Answer:
No. It cannot be negative because the minimum potential energy of a particle executing simple harmonic motion at mean position is zero. The potential energy increases in positive direction at the extreme position.
However, if we choose zero potential energy at some other point, say extreme position, the potential energy can be negative at the mean position.
Question 10:
The energy of system in simple harmonic motion is given by
E=12m ω2A2.Which of the following two statements is more appropriate?
(A) The energy is increased because the amplitude is increased.
(B) The amplitude is increased because the energy is increased.
Answer:
Statement A is more appropriate because the energy of a system in simple harmonic motion is given by
E=12m ω2A2.If the mass (m) and angular frequency (ω) are made constant, Energy (E) becomes proportional to the square of amplitude (A^{2}).
i.e. E ∝ A^{2}
Therefore, according to the relation, energy increases as the amplitude increases.
Question 11:
A pendulum clock gives correct time at the equator. Will it gain time or loose time as it is taken to the poles?
Answer:
According to the relation:
T=2πlgThe time period (T) of the pendulum becomes proportional to the square root of inverse of g if the length of the pendulum is kept constant.
i.e.
T∝1gAlso, acceleration due to gravity (g) at the poles is more than that at equator. Therefore, the time period decreases and the clock gains time.
Question 12:
Can a pendulum clock be used in an earthsatellite?
Answer:
No. According to the relation:
T=2πlgThe time period of the pendulum clock depends upon the acceleration due to gravity. As the earthsatellite is a free falling body and its g_{effective} (effective acceleration due to gravity ) is zero at the satellite, the time period of the clock is infinite.
Question 13:
A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simple pendulum is valid with the distance between the point of suspension and centre of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?
Answer:
The time period of a pendulum depends on the length and is given by the relation,
T=2πlg.
As the effective length of the pendulum first increases and then decreases, the time period first increases and then decreases.
Question 14:
A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?
Answer:
Yes.
Time period of a spring mass system is given by,
T=2πmk …(1) .
where m is mass of the block, and
k is the spring constant
Time period is also given by the relation,
T=2πx0g …(2)
where, x_{0} is extension of the spring, and
g is acceleration due to gravity
From the equations (1) and (2), we have:
mg=kx0 ⇒k=mgx0Substituting the value of k in the above equation, we get:
T=2πmmgx0=2πx0gThus, we can find the time period if the value of extension x_{0} is known.
Question 15:
A platoon of soldiers marches on a road in steps according to the sound of a marching band. The band is stopped and the soldiers are ordered to break the steps while crossing a bridge. Why?
Answer:
When the frequency of soldiers’ feet movement becomes equal to the natural frequency of the bridge, and resonance occurs between soldiers’ feet movement and movement of the bridge, maximum transfer of energy occurs from soldiers’ feet to the bridge, which increases the amplitude of vibration. A continued increase in the amplitude of vibration, however, may lead to collapsing of the bridge.
Question 16:
The force acting on a particle moving along Xaxis is F = −k(x − v_{o} t) where k is a positive constant. An observer moving at a constant velocity v_{0} along the Xaxis looks at the particle. What kind of motion does he find for the particle?
Answer:
As the observer moves with a constant velocity along the same axis, he sees the same force on the particle and finds the motion of the particle is not simple harmonic motion.
Question 1:
A student says that he had applied a force
F=kxon a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.
(a) As x increases k increases.
(b) As x increases k decreases.
(c) As x increases k remains constant.
(d) The motion cannot be simple harmonic.
Answer:
(a) As x increases k increases.
A body is said to be in simple harmonic motion only when,
F =
–kx …(1)
where F is force,
k is force constant, and
x is displacement of the body from the mean position.
Given:
F = –k
x …(2)
On comparing the equations (1) and (2), it can be said that in order to execute simple harmonic motion, k should be proportional to
x. Thus, as x increases k increases.
Question 2:
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is
(a) the mean position
(b) an extreme position
(c) between the mean position and the positive extreme
(d) between the mean position and the negative extreme
Answer:
(b) an extreme position
One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started.
Question 3:
The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity
v→. The value of v is
(a) v_{max}
(b) 0
(c) between 0 and v_{max}
(d) between 0 and −v_{max}
Answer:
(a) v_{max}
Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation_{.}
Question 4:
The displacement of a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d) zero
Answer:
(d) zero
Displacement is defined as the distance between the starting and the end point through a straight line. In one complete oscillation, the net displacement is zero as the particle returns to its initial position.
Question 5:
The distance moved by a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d) zero
Answer:
(c) 4A
In an oscillation, the particle goes from one extreme position to other extreme position that lies on the other side of mean position and then returns back to the initial extreme position. Thus, total distance moved by particle is,
2A + 2A = 4A.
Question 6:
The average acceleration in one time period in a simple harmonic motion is
(a) Aω^{2}
(b) Aω^{2}/2
(c)
Aω2/2(d) zero
Answer:
(d) zero
The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
Aω^{2} + ( –Aω^{2}) = 0
Question 7:
The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A + B
(c) simple harmonic with amplitude (A + B)/2
(d) simple harmonic with amplitude
A2+B2.
Answer:
(d) simple harmonic with amplitude
A2+B2
x = A sin ωt + B cos ωt …(1)
Acceleration, a=d2xdt2=d2dt2(A sin ωt + B cos ωt) =ddt(Aω cos ωt – Bω sin ωt) =Aω2 sinωt – Bω2 cos ωt =ω2(A sin ωt + B cos ωt ) = ω2x
For a body to undergo simple harmonic motion,
acceleration, a =
–kx. …(2)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude, A
=A2+B2.2
Question 8:
The displacement of a particle is given by
r→=Ai→ cosωt + j→sinωt.The motion of the particle is
(a) simple harmonic
(b) on a straight line
(c) on a circle
(d) with constant acceleration
Answer:
(c) on a circle
We know,
d2dt2r→=ω2 r→But there is a phase difference of 90^{o} between the x and y components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.
Question 9:
A particle moves on the Xaxis according to the equation x = A + B sin ωt. The motion is simple harmonic with amplitude
(a) A
(b) B
(c) A + B
(d)
A2+B2.
Answer:
(b) B
At t = 0,
Displacement
x0is given by,
x_{0} = A + sin ω(0) = A
Displacement x will be maximum when sinωt is 1
or,
x_{m} = A + B
Amplitude will be:
x_{m}
–x_{o} = A + B
A = B
Question 10:
Figure represents two simple harmonic motions.
Figure
The parameter which has different values in the two motions is
(a) amplitude
(b) frequency
(c) phase
(d) maximum velocity
Answer:
(c) phase
Because the direction of motion of particles A and B is just opposite to each other.
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Question 11:
The total mechanical energy of a springmass system in simple harmonic motion is
E=12mω2A2.Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will
(a) become 2E
(b) become E/2
(c) become
2E(d) remain E
Answer:
(d) remain E
Mechanical energy (E) of a springmass system in simple harmonic motion is given by,
E=12mω2A2where m is mass of body, and
ωis angular frequency.
Let m_{1} be the mass of the other particle and ω_{1} be its angular frequency.
New angular frequency ω_{1} is given by,
ω1=km1=k2m (m1=2m)New energy E_{1} is given as,
E1 = 12m1ω12A2 =12(2m)(k2m)2A2 =12mω2A2 =E
Question 12:
The average energy in one time period in simple harmonic motion is
(a)
12m ω2A2(b)
14m ω2A2(c) m ω^{2}A^{2}
(d) zero
Answer:
(a)
12mω2A2It is the total energy in simple harmonic motion in one time period.
Question 13:
A particle executes simple harmonic motion with a frequency v. The frequency with which the kinetic energy oscillates is
(a) v/2
(b) v
(c) 2 v
(d) zero
Answer:
(c) 2v
Because in one complete oscillation, the kinetic energy changes its value from zero to maximum, twice.
Question 14:
A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is T. If the spring is divided in two equal parts and one part is used to continue the simple harmonic motion, the time period will
(a) remain T
(b) become 2T
(c) become T/2
(d) become
T/2
Answer:
(d) become T/
2T/2√
Time period
Tis given by,
T = 2πmkwhere m is the mass, and
k is spring constant.
When the spring is divided into two parts, the new spring constant k_{1} is given as,
k_{1} =
2kNew time period T_{1}:
T_{1 }=
2πm2k=122πmk=12T
Question 15:
Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k_{1} and k_{2} respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
(a) k_{1}/k_{2}
(b)
k1/k2(c) k_{2}/k_{1}
(d)
k2/k1
Answer:
(d)
k2k1Maximum velocity, v = Aω
where A is amplitude and ω is the angular frequency.
Further, ω =
kmLet A and B be the amplitudes of particles A and B respectively. As the maximum velocity of particles are equal,
i.e. vA=vBor, AωA=BωB⇒ Ak1mA= Bk2mB⇒ Ak1m= Bk2m (mA=mB=m)⇒ AB= k2k1
Question 16:
A springmass system oscillates with a frequency v. If it is taken in an elevator slowly accelerating upward, the frequency will
(a) increase
(b) decrease
(c) remain same
(d) become zero
Answer:
(c) remain same
Because the frequency (
ν=12πkm) of the system is independent of the acceleration of the system.
Question 17:
A springmass system oscillates with a car. If the car accelerates on a horizontal road, the frequency of oscillation will
(a) increase
(b) decrease
(c) remain same
(d) become zero
Answer:
(c) remain same
As the frequency of the system is independent of the acceleration of the system.
Question 18:
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run
(a) at correct rate
(b) 6 times faster
(c)
6times faster
(d)
6times slower
Answer:
(d)
6times slower
The acceleration due to gravity at moon is g/6.
Time period of pendulum is given by,
T = 2πlgTherefore, on moon, time period will be:
T_{moon} =
2πlgmoon=2πl(g6)=6(2πlg)=6T
Question 19:
A wall clock uses a vertical springmass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives correct time at the equator. If the clock is taken to the poles it will
(a) run slow
(b) run fast
(c) stop working
(d) give correct time
Answer:
(d) give correct time
Because the time period of a springmass system does not depend on the acceleration due to gravity.
Question 20:
A pendulum clock keeping correct time is taken to high altitudes,
(a) it will keep correct time
(b) its length should be increased to keep correct time
(c) its length should be decreased to keep correct time
(d) it cannot keep correct time even if the length is changed
Answer:
(c) its length should be decreased to keep correct time
Time period of pendulum,
T =
2πlgAt higher altitudes, the value of acceleration due to gravity decreases. Therefore, the length of the pendulum should be decreased to compensate for the decrease in the value of acceleration due to gravity.
Question 21:
The free end of a simple pendulum is attached to the ceiling of a box. The box is taken to a height and the pendulum is oscillated. When the bob is at its lowest point, the box is released to fall freely. As seen from the box during this period, the bob will
(a) continue its oscillation as before
(b) stop
(c) will go in a circular path
(d) move on a straight line
Answer:
(c) will go in a circular path
As the acceleration due to gravity acting on the bob of pendulum, due to free fall gives a torque to the pendulum, the bob goes in a circular path.
Question 1:
Select the correct statements.
(a) A simple harmonic motion is necessarily periodic.
(b) A simple harmonic motion is necessarily oscillatory.
(c) An oscillatory motion is necessarily periodic.
(d) A periodic motion is necessarily oscillatory.
Answer:
(a) A simple harmonic motion is necessarily periodic.
(b) A simple harmonic motion is necessarily oscillatory.
A periodic motion need not be necessarily oscillatory. For example, the moon revolving around the earth.
Also, an oscillatory motion need not be necessarily periodic. For example, damped harmonic motion.
Question 2:
A particle moves in a circular path with a uniform speed. Its motion is
(a) periodic
(b) oscillatory
(c) simple harmonic
(d) angular simple harmonic
Answer:
(a) periodic
Because the particle covers one rotation after a fixed interval of time but does not oscillate around a mean position.
Question 3:
A particle is fastened at the end of a string and is whirled in a vertical circle with the other end of the string being fixed. The motion of the particle is
(a) periodic
(b) oscillatory
(c) simple harmonic
(d) angular simple harmonic
Answer:
(a) periodic
Because the particle completes one rotation in a fixed interval of time but does not oscillate around a mean position.
Question 4:
A particle moves in a circular path with a continuously increasing speed. Its motion is
(a) periodic
(b) oscillatory
(c) simple harmonic
(d) none of them
Answer:
(d) none of them
As the particle does not complete one rotation in a fixed interval of time, neither does it oscillate around a mean position.
Question 5:
The motion of a torsional pendulum is
(a) periodic
(b) oscillatory
(c) simple harmonic
(d) angular simple harmonic
Answer:
(a) periodic
(b) oscillatory
(d) angular simple harmonic
Because it completes one oscillation in a fixed interval of time and the oscillations are in terms of rotation of the body through some angle.
Question 6:
Which of the following quantities are always negative in a simple harmonic motion?
(a)
F→. a→.(b)
v→. r→.(c)
a→. r→.(d)
F→. r→.
Answer:
(c) a⇀.r⇀(d) F⇀.r⇀In S.H.M.,
F = –kx
Therefore,
F.⇀r⇀will always be negative. As acceleration has the same direction as the force,
a⇀.r⇀will also be negative, always.
Question 7:
Which of the following quantities are always positive in a simple harmonic motion?
(a)
F→. a→.(b)
v→. r→.(c)
a→. r→.(d)
F→. r→.
Answer:
(a) F⇀.a⇀As the direction of force and acceleration are always same,
F⇀.a⇀is always positive.
Question 8:
Which of the following quantities are always zero in a simple harmonic motion?
(a)
F→× a→.(b)
v→× r→.(c)
a→× r→.(d)
F→× r→.
Answer:
(a)F⇀ x a⇀(b)v⇀ x r⇀⇀(c)a⇀x r⇀(d)F⇀ x r⇀As
F⇀, a⇀, r⇀ and v⇀are either parallel or antiparallel to each other, their cross products will always be zero.
Question 9:
Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance h directly above the tunnel. The motion of the particle as seen from the earth is
(a) simple harmonic
(b) parabolic
(c) on a straight line
(d) periodic
Answer:
(c) on a straight line
(d) periodic
If the particle were dropped from the surface of the earth, the motion of the particle would be SHM. But when it is dropped from a height h, the motion of the particle is not SHM because there is no horizontal velocity imparted. In that case, the motion of the particle would be periodic and in a straight line.
Page No 252:
Question 1:
A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm going towards positive xdirection. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s.
Answer:
It is given,
Amplitude of the simple harmonic motion, A =10 cm
At t = 0 and x = 5 cm,
Time period of the simple harmonic motion, T = 6 s
Angular frequency (ω) is given by,
ω=2πT=2π6=π3 sec1Consider the equation of motion of S.H.M,
Y = Asin
ωt+ϕ …(1)
where Y is displacement of the particle, and
ϕis phase of the particle.
On substituting the values of A, t and ω in equation (1), we get:
5 = 10sin(ω × 0 + Ï•)
⇒5 = 10sin Ï•
sin ϕ=12⇒ϕ=π6∴ Equation of displacement can be written as,
x=10 cm sin π3t+π6(ii) At t = 4 s,
x=10sinπ34+π6 =10sin8π+π6 =10sin9π6 =10sin3π2 = 10sinπ+π2 =10sinπ2=10Acceleration is given by,
a = −ω^{2}x
=π29×10=10.9≈11 cm/sec2
Question 2:
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitude 2 cm, 1 m s^{−1} and 10 m s^{−2} at a certain instant. Find the amplitude and the time period of the motion.
Answer:
It is given that:
Position of the particle, x = 2 cm = 0.02 m
Velocity of the particle, v = 1 ms^{−1}^{.}
Acceleration of the particle, a = 10 ms^{−2}^{.}
Let
ω be the angular frequency of the particle.
The acceleration of the particle is given by,
a = ω^{2}x
⇒ω=ax=100.02 =500=105 HzTime period of the motion is given as, T=2πω=2π105 =2×3.1410×2.236 =0.28 sNow, the amplitude A is calculated as,
v=ωA2x2 ⇒v2=ω2 A2x2 1=500A20.0004 ⇒A=0.0489=0.049 m ⇒A= 4.9 cm
Question 3:
A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?
Answer:
It is given that:
Amplitude of the particle executing simple harmonic motion, A = 10 cm
To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let y be displacement of the particle,
ω be the angular speed of the particle, and
A be the amplitude of the simple harmonic motion.
Equating the mathematical expressions for K.E. and P.E. of the particle, we get:
12mω2 A2y2=12mω2y2 A^{2} − y^{2} = y^{2}
2y^{2} = A^{2}
⇒y=A2=102=52The kinetic energy and potential energy of the particle are equal at a distance of
52 cm from the mean position.
Question 4:
The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s^{−1} and 50 cm s^{−2}. Find the position(s) of the particle when the speed is 8 cm s^{−1}.
Answer:
It is given that:
Maximum speed of the particle,
vMax= 10 cms
1
Maximum acceleration of the particle,
aMax= 50 cms^{−2}â€‹
The maximum velocity of a particle executing simple harmonic motion is given by,
vMax=Aωwhere
ω is angular frequency, and A is amplitude of the particle.
Substituting the value of
vMaxin the above expression, we get:
Aω = 10
…(1)
⇒ω2=100A2 a_{Max} = ω^{2}A = 50 cms^{−1}
⇒ω2=50A …(2)From the equations (1) and (2), we get:100A2=50A ⇒A=2 cm∴ ω=100A2=5 sec1To determine the positions where the speed of the particle is 8 ms^{1}, we may use the following formula:
v^{2} = ω^{2} (A^{2} − y^{2})
where y is distance of particle from the mean position, and
v is velocity of the particle.
On substituting the given values in the above equation, we get:
64 = 25 (4 − y^{2})
⇒6425=4y2⇒ 4 − y^{2} = 2.56
⇒ y^{2} = 1.44
⇒â€‹ y =
1.44⇒ y = ± 1.2 cm (from the mean position)
Question 5:
A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [(100 s^{−1})t + π/6]. Find (a) the amplitude, the time period and the spring constant. (c) the position, the velocity and the acceleration at t = 0.
Answer:
Given:
Equation of motion of the particle executing S.H.M.,
x=2.0 cm sin 100 s1t+π6Mass of the particle, m=10 g …(1)
General equation of the particle is given by,
x=Asin(ωt+ϕ) …(2)
On comparing the equations (1) and (2) we get:
(a) Amplitude, A is 2 cm.
Angular frequency, ω is 100 s^{−1}â€‹.
Time period is calculated as, T=2πω=2π100=π50s =0.063 sAlso, we know –
T=2πmkwhere k is the spring constant.⇒T2=4π2mk⇒ k=4π2mT2=105 dyne/cm =100 N/m(b) At t = 0 and x = 2 cm
sinπ6
=2×12=1 cm from the mean position, We know:
x = A sin (ωt + Ï•)
Using
v=dxdt,we get:
v = Aω cos (ωt + Ï•)
=2×100 cos 0+π6=200×32=1003 cms1=1.73 ms1(c) Acceleration of the particle is given by,
a =
ω2x
= 100^{2}×1 = 10000 cm/s^{2}
Question 6:
The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + π/3), where x is in centimetre and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed?
Answer:
Given:
The equation of motion of a particle executing S.H.M. is,
x=5 sin 20t+π3
The general equation of S..H.M. is give by,
x=A sin (ωt+ϕ)
(a) Maximum displacement from the mean position is equal to the amplitude of the particle.
As the velocity of the particle is zero at extreme position, it is at rest.
∴ Displacementx = 5, which is also the amplitude of the particle.
⇒5=5 sin 20t+π3Now, sin 20t+π3=1=sinπ2 ⇒ 20t+π3=π2 ⇒ t=π120 sThe particle will come to rest at
π120 s(b) Acceleration is given as,
a = ω^{2}x
=ω25 sin 20t+π3
For a = 0,
5 sin 20t+π3=0⇒sin 20t+π3=sin π⇒20t=ππ3=2π3⇒ t=π30 s(c) The maximum speed
vis given by,
v=Aωcos ωt+π3 (using
v=dxdt)
=20×5 cos 20t+π3 For maximum velocity:
cos 20t+π3=1= cos π⇒20t=ππ3=2π3⇒ t=π30 s
Question 7:
Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 πt + tan^{−1} 0.75) where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the first time? (b) When does he acceleration have its maximum magnitude for the first time? (c) When does the particle come to rest for
Answer:
It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
x = 2.0 cos (50
πt + tan^{−1}0.75)
= 2.0 cos (50
πt + 0.643)
(a) Velocity of the particle is given by,
v=dxdt v = −100
πsin (50
πt + 0.643)
As the particle comes to rest, its velocity becomes be zero.
⇒â€‹ v = −100
πsin (50
πt + 0.643) = 0
⇒ sin (50
πt + 0.643) = 0 = sin
π When the particle initially comes to rest,
50
πt + 0.643 =
π ⇒ t = 1.6 × 10^{−2} s
(b) Acceleration is given by,
a=dvdt =100π×50π cos 50πt+0.643 For maximum acceleration:
cos (50
πt + 0.643) = −1 = cos
π(max) (so that a is max)
⇒ t = 1.6 × 10^{−2} s
(c) When the particle comes to rest for the second time, the time is given as,
50
πt + 0.643 = 2
π⇒ â€‹ t = 3.6 × 10^{−2} s
Question 8:
Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from half the amplitude to the amplitude.
Answer:
As per the conditions given in the question,
y1=A2;y2=A (for the given two positions)
Let y_{1} and y_{2} be the displacements at the two positions and A be the amplitude.
Equation of motion for the displacement at the first position is given by,
y_{1} = Asinωt_{1}
As displacement is equal to the half of the amplitude,
A2=A sin ωt1
⇒ sin ωt1=12⇒2π×t1T=π6⇒ t1=T12The displacement at second position is given by,
y_{2} = A sin ωt_{2}
As displacement is equal to the amplitude at this position,
⇒ A = A sin ωt_{2}
⇒ sinωt_{2} = 1
⇒ ωt2=π2⇒2πTt2=π2 ∵ sin π2=1⇒ t2=T4∴ t2t1=T4T12=T6
Question 9:
The pendulum of a clock is replaced by a springmass system with the spring having spring constant 0.1 N m^{−1}. What mass should be attached to the spring?
Answer:
Given:
Spring constant, k =0.1 N/m
Time period of the pendulum of clock, T = 2 s
Mass attached to the string, m, is to be found.
The relation between time period and spring constant is given as,
T=2π mkOn substituting the respective values, we get:
2=2πmk⇒ π2m0.1=1∴ m=0.1π2=0.110 =0.01 kg≈10 g
Question 10:
A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum, i.e., a pendulum having frequency same as that of the block.
Answer:
An equivalent simple pendulum has same time period as that of the spring mass system.
The time period of a simple pendulum is given by,
Tp=2πlgwhere l is the length of the pendulum, and
g is acceleration due to gravity.
Time period of the spring is given by,
Ts=2πmkwhere m is the mass, and
k is the spring constant.
Let x be the extension of the spring.
For small frequency, T_{P} â€‹can be taken as equal to T_{S.}
⇒lg=mk⇒lg=mk⇒l=mgk=Fk=x(
∵restoring force = weight = mg)
∴l = x (proved)
Question 11:
A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.
Answer:
It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Time period of simple harmonic motion, T = 0.314 s
Mass of the block, m = 0.5 kg
Weight of the block, W = mg = 0.5
×10 = 5 kg
∵ g=10 ms2Total force exerted on the block = Weight of the block + spring force
Periodic time of spring is given by,
T=2πmk⇒0.314=2π0.5k⇒ k=200 N/m∴ The force exerted by the spring on the block
Fis,
F = kx = 200.0 × 0.1 = 20 N
Maximum force = F + weight of the block
= 20 + 5 = 25 N
Question 12:
A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring.
Answer:
It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for springmass system is given as,
T=2πmkwhere k is the spring constant.
On substituting the respective values, we get:
⇒4=2π2k⇒2=π2k⇒4=π22k⇒k=2π24 =π22= 5 N/mAs the restoring force is balanced by the weight, we can write:
F = mg = kx
⇒x=mgk=2×105=4∴ Potential Energy
Uof the spring is,
U=12kx2=12×5×16 =5×8=40 J
Question 10:
For a particle executing simple harmonic motion, the acceleration is proportional to
(a) displacement from the mean position
(b) distance from the mean position
(c) distance travelled since t = 0
(d) speed
Answer:
(a) displacement from the mean position
For S.H.M.,
F = kx
ma = – kx (F = ma)
or,
a =
kmxThus, acceleration is proportional to the displacement from the mean position but in opposite direction.
Question 11:
A particle moves in the X–Y plane according to the equation
r→=i→+2j→Acosωt.The motion of the particle is
(a) on a straight line
(b) on an ellipse
(c) periodic
(d) simple harmonic
Answer:
(a) on a straight line
(c) periodic
(d) simple harmonic
The given equation is a solution to the equation of simple harmonic motion. The amplitude is
(i⇀+2j⇀)A, following equation of straight line y = mx + c. Also, a simple harmonic motion is periodic.
Question 12:
A particle moves on the Xaxis according to the equation x = x_{0} sin^{2} ωt. The motion is simple harmonic
(a) with amplitude x_{0}
(b) with amplitude 2x_{0}
(c) with time period
2πω(d) with time period
πω.
Answer:
(d) with time period
πωGiven equation:
x = x_{o} sin^{2} ωt
⇒â€‹
x=x02(cos 2ωt 1)Now, the amplitude of the particle is x_{o}/2 and the angular frequency of the SHM is 2ω.
Thus, time period of the SHM =
2πangular frequency=2π2ω=πω
Question 13:
In a simple harmonic motion
(a) the potential energy is always equal to the kinetic energy
(b) the potential energy is never equal to the kinetic energy
(c) the average potential energy in any time interval is equal to the average kinetic energy in that time interval
(d) the average potential energy in one time period is equal to the average kinetic energy in this period.
Answer:
(d) the average potential energy in one time period is equal to the average kinetic energy in this period.
The kinetic energy of the motion is given as,
12kA2 cos2 ωt
The potential energy is calculated as,
12kA2 sin2 ωtAs the average of the cosine and the sine function is equal to each other over the total time period of the functions, the average potential energy in one time period is equal to the average kinetic energy in this period.
Question 14:
In a simple harmonic motion
(a) the maximum potential energy equals the maximum kinetic energy
(b) the minimum potential energy equals the minimum kinetic energy
(c) the minimum potential energy equals the maximum kinetic energy
(d) the maximum potential energy equals the minimum kinetic energy
Answer:
(a) the maximum potential energy equals the maximum kinetic energy
(b) the minimum potential energy equals the minimum kinetic energy
In SHM,
maximum kinetic energy =
12kA2
maximum potential energy =
12kA2The minimum value of both kinetic and potential energy is zero.
Therefore, in a simple harmonic motion the maximum kinetic energy and maximum potential energy are equal. Also, the minimum kinetic energy and the minimum potential energy are equal.
Question 15:
An object is released from rest. The time it takes to fall through a distance h and the speed of the object as it falls through this distance are measured with a pendulum clock. The entire apparatus is taken on the moon and the experiment is repeated
(a) the measured times are same
(b) the measured speeds are same
(c) the actual times in the fall are equal
(d) the actual speeds are equal
Answer:
(a) the measured times are same
(b) the measured speeds are same
The effect of gravity on the object as well as on the pendulum clock is same in both cases; the time measured is also same. As the time measured is same, the speeds are same.
Question 16:
Which of the following will change the time period as they are taken to moon?
(a) A simple pendulum
(b) A physical pendulum
(c) A torsional pendulum
(d) A springmass system
Answer:
(a) A simple pendulum
(b) A physical pendulum
As the time period of a simple pendulum and a physical pendulum depends on the acceleration due the gravity, the time period of these pendulums changes when they are taken to the moon.
Page No 253:
Question 13:
A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block?
Answer:
It is given that:
Energy stored in the spring, E = 5 J
Frequency of the massspring system, f = 5
Extension in the length of the spring, x = 25 cm = 0.25 m
Time period, T =15 sPotential energyU is given by, U=12kx2⇒12kx2=5⇒12k0.252=5⇒k=160 N/mTime period of spring mass system is given by, T=2πmk where m is the mass of the body hanged, and k is the spring constant.On substituting the respective values in the above expression, we get: 15=2πm160⇒m=0.16 kg
Question 14:
A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude? (c) What can be the maximum amplitude with which the two blocks may oscillate together?
Figure
Answer:
(a) Consider the free body diagram.
Weight of the body, W = mg
Force, F = ma = mω^{2}x
x is the small displacement of mass m.
As normal reaction R is acting vertically in the upward direction, we can write:
R + mω^{2}x − mg = 0 ….(1)
Resultant force = mω^{2}x = mg − R
⇒mω2x=mkM+mx =mkxM+mHere,ω=kM+m(b) R = mg − mω^{2}x
=mgmkM+Nx=mgmkxM+NIt can be seen from the above equations that, for R to be smallest, the value of mω^{2}x should be maximum which is only possible when the particle is at the highest point.
(c) R = mg − mω^{2}x
As the two blocks oscillate together R becomes greater than zero.
When limiting condition follows,
i.e. R = 0
mg = mω^{2}x
x=mgmω2=mg·M+mmk Required maximum amplitude
=gM+mk
Question 15:
The block of mass m_{1} shown in figure (12−E2) is fastened to the spring and the block of mass m_{2} is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k) (m_{1} + m_{2})g sin θ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?
Figure
Answer:
(a) As it can be seen from the figure,
Restoring force = kx
Component of total weight of the two bodies acting vertically downwards = (m_{1} + m_{2}) g sin θ
At equilibrium,
kx = (m_{1} + m_{2}) g sin θ
⇒x=m1+m2 g sin θk
(b) It is given that:
Distance at which the spring is pushed,
x1=2km1+m2g sin θAs the system is released, it executes S.H.M.
where
ω=km1+m2When the blocks lose contact, P becomes zero. (P is the force exerted by mass m_{1} on mass m_{2})
∴ m2g sin θ=m2x2ω2=m2x2×km1+m2⇒x2=m1+m2 g sin θkTherefore, the blocks lose contact with each other when the spring attains its natural length.
(c) Let v be the common speed attained by both the blocks.
Total compression = x1+x212 m1+m2v20 = 12kx1+x22m1+m2 g sin θ x+x1 ⇒12m1+m2v2=12k3k m1+m2 g sin θm1+m2 g sin θ x1+x2⇒12m1+m2v2=12 m1+m2 g sin θ×3k m1+m2 g sin θ⇒v=3k m1+m2 g sin θ
Page No 254:
Question 28:
The left block in figure (12−E13) moves at a speed v towards the right block placed in equilibrium. All collisions to take place are elastic and the surfaces are frictionless. Show that the motions of the two blocks are periodic. Find the time period of these periodic motions. Neglect the widths of the blocks.
Figure
Answer:
According to the question, the collision is elastic and the surface is frictionless, therefore, when the left block A moves with speed v and collides with the right block B, it transfers all the energy to the right block B.
The left block A moves a distance x against the spring; the right block returns to the original position and completes half of the oscillation.
Therefore, the period of right block B will be,
T=2πmk2=πmkRight block B collides with left block A and comes to rest.
Let L be the distance moved by the block to return to its original position.
The time taken is given by,
LV+LV=2LVHence, time period of the periodic motion is,
2LV+πmk.
Question 29:
Find the time period of the motion of the particle shown in figure (12−E14). Neglect the small effect of the bend near the bottom.
Figure
Answer:
Let t_{1} and t_{2} be the time taken by the particle to travel distances AB and BC respectively.
Acceleration for part AB, a_{1}_{ =} g sin 45°
The distance travelled along AB is s_{1}_{.}
∴ s1=0.1sin 45°=2 mLet v be the velocity at point B, and
u be the initial velocity.
Using the third equation of motion, we have:
v^{2} − u^{2} = 2a_{1}s_{1}
⇒v2=2×g sin 45°×0.1sin 45°=2⇒v=2 m/sAs v=u+a1t1∴t1=vua1 =20g2 =2g=210=0.2 sec ( g=10 ms2)For the distance BC,
Acceleration, a_{2} =
–gsin 60°
Initial velocity, u=2 v = 0∴ time period, t2=02g32=223g =2×1.4141.732×10=0.163 sThus, the total time period, t = 2(t_{1} + t_{2}) = 2 (0.2 + 0.163) = 0.73 s
Question 30:
All the surfaces shown in figure (12−E15) are frictionless. The mass of the care is M, that of the block is m and the spring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through x length x_{0} when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and of the care as seen from the road. (b) Find the time period(s) of the two simple harmonic motions.
Figure
Answer:
Let x_{1} and x_{2} be the amplitudes of oscillation of masses m and M respectively.
(a) As the centre of mass should not change during the motion, we can write:
mx_{1} = Mx_{2}
…(1)Let k be the spring constant. By conservation of energy, we have:
12kx02=12kx1+x22 …(2) where x_{0} is the length to which spring is stretched.
From equation (2) we have,
x0=x1+x2On substituting the value of x_{2} from equation (1) in equation (2), we get:
x0=x1+mx1M⇒x0=1+mMx1⇒x1=MM+mx0
Now, x2=x0x1On substituting the value of x_{1} from above equation, we get:
⇒ x2=x01MM+m⇒ x2=mx0M+mThus, the amplitude of the simple harmonic motion of a car, as seen from the road is
mx0M+m.
(b) At any position,
Let v_{1} and v_{2} be the velocities.
Using law of conservation of energy we have,
12Mv2+12mv1v22+12kx1+x22=constant …3Here, (v_{1} − v_{2}) is the absolute velocity of mass m as seen from the road.
Now, from the principle of conservation of momentum, we have:
Mx_{2} = mx_{1}
⇒x1=Mmx2 ….4Mv2=mv1v2⇒v1v2=Mmv2 ….5Putting the above values in equation (3), we get:
12Mv22+12mM2m2v22+12kx221+Mm2=constant∴M1+Mmv22+k1+Mmx22=constant⇒Mv22+k1+Mmx22=constantTaking derivative of both the sides, we get:
M×2v2dv2dt+kM+mm2x2dx2dt=0⇒ma2+kM+mmx2=0 because, v2=dx2dta2x2=kM+mMm=ω2∴ω=kM+mMmTherefore, time period, T=2πMmkM+m
Question 31:
A uniform plate of mass M stays horizontally and symmetrically on two wheels rotating in opposite direction (figure 12−E16). The separation between the wheels is L. The friction coefficient between each wheel and the plate is μ. Find the time period of oscillation of the plate if it is slightly displaced along its length and released.
Figure
Answer:
Let x be the displacement of the uniform plate towards left.
Therefore, the centre of gravity will also be displaced through displacement x.
At the displaced position,
R_{1} + R_{2} = mg
Taking moment about g, we get:
R1L2x=R2L2+x=MgR1 L2+x ….1∴ R1L2x=MgR1 L2+x⇒R1L2R1x=MgL2R1x+MgxR1L2⇒R1L2+R1L2=Mgx+L2⇒R1L2+L2=Mg2x+L2⇒R1L=Mg2x+L⇒R1=MgL+2x2LNow, F1=μR1=μMgL+2x2LSimilarly, F2=μR2=μMgL2x2LAs F1>F2, we can write:F1F2=Ma=2μMgLx ax=2μgL=ω2⇒ω=2μgLTime periodT is given by,T=2πL2μg
Question 32:
A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a second pendulum at a place where g = π^{2} m s^{−2}.
Answer:
It is given that:
Time period of the second pendulum, T = 2 s
Acceleration due to gravity of a given place, g =
π^{2} ms^{−2}
The relation between time period and acceleration due to gravity is given by,
T=2πlgwhere l is the length of the second pendulum.
Substituting the values of T and g, we get:
⇒2=2πlπ2⇒1π=lπ⇒l=1 mHence, the length of the pendulum is 1 m.
Question 33:
The angle made by the string of a simple pendulum with the vertical depends on time as
θ=π90 sin π s1t. Find the length of the pendulum if g = π^{2} m^{−2}.
Answer:
It is given that:Angle made by the simple pendulum with the vertical, θ=π90sin πs1tOn comparing the above equation with the equation of S.H.M., we get: ω= π s1 ⇒2πT=π∴ T=2 sTime period is given by the relation,T=2πlg⇒2=2πlπ2⇒1=π1πl⇒l=1 mHence, length of the pendulum is 1 m.
Page No 255:
Question 34:
The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours?
Answer:
Given,
Time period of the clock pendulum = 2.04 s
The number of oscillations made by the pendulum in one day is calculated as
Nmber of seconds in one daytime period of pendulum in seconds=24×36002= 43200
In each oscillation, the clock gets slower by (2.04 − 2.00) s, i.e., 0.04 s.
In one day, it is slowed by = 43200 × (0.04)
= 28.8 min
Thus, the clock runs 28.8 minutes slow during 24 hours.
Question 35:
A pendulum clock giving correct time at a place where g = 9.800 m s^{−2} is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.
Answer:
Let T_{1} be the time period of pendulum clock at a place where acceleration due to gravity
g1is 9.8 ms^{−2}.
Let T_{1 = }2 s
g_{1} = 9.8 ms
2
Let T_{2} be the time period at the place where the pendulum clock loses 24 seconds during 24 hours.
Acceleration due to gravity at this place is
g2.
T2=24×360024×3600242 =2×36003599As
T∝1g
∴T1T2=g2g1⇒ g2g1=T1T22⇒g2=9.8 359936002 =9.795 m/s2
Question 36:
A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is 1.67 m s^{−2}?
Answer:
It is given that:
Length of the pendulum, l = 5 m
Acceleration due to gravity, g = 9.8 ms^{2}
Acceleration due to gravity at the moon, g‘ = 1.67 ms^{2}
(a) Time period
Tis given by,
T=2πlg
=2π59.8=2π0.510=2π 0.71 si.e. the body will take 2
π(0.7) seconds to complete an oscillation.
Now, frequency
fis given by,
f=1T
∴ f=12π0.71 =0.70π Hz(b) Let
g’be the value of acceleration due to gravity at moon. Time period of simple pendulum at moon
T’, is given as:
T’=2πlg’On substituting the respective values in the above formula, we get:
T’=2π51.67Therefore, frequency
f’will be,
f’=1T’ =12π1.675=12π0.577 =12π3 Hz
Question 37:
The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.
Answer:
Let the speed of bob of the pendulum at an angle
θbe v.
Using the principle of conservation of energy between the mean and extreme positions, we get:
12mv^{2} − 0 = mgl(1 − cos θ)
v^{2} = 2gl(1 − cos θ) …(1)
In a moving pendulum, the tension is maximum at the mean position, whereas it is minimum at the extreme position.
Maximum tension at the mean position is given by
T_{max} = mg + 2mg(1 − cos θ)
Minimum tension at the extreme position is given by
T_{min} = m g cosθ
According to the question,
T_{max} = 2T_{min}
⇒ mg + 2mg − 2m g cosθ = 2m g cosθ
⇒ 3mg = 4 mg cosθ
⇒ cos θ=34⇒θ=cos1 34
Question 38:
A small block oscillates back and forth on a smooth concave surface of radius R (figure 12−E17). Find the time period of small oscillation.
Figure
Answer:
It is given that R is the radius of the concave surface.
â€‹Let N be the normal reaction force.
Driving force, F = mg sin θ
Comparing the expression for driving force with the expression, F = ma, we get:
Acceleration, a = g sin θ
Since the value of θ is very small,
∴ sin θ → θ
∴ Acceleration, a = gθ
Let x be the displacement of the body from mean position.
∴θ=xR⇒a=gθ=gxR⇒ax=gR
⇒a=xgRAs acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.
Time period
Tis given by,
T=2πdisplacementAcceleration
=2πxgx/R=2πRg
Question 39:
A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period.
Answer:
Let ω be the angular velocity of the system about the point of suspension at any time.
Velocity of the ball rolling on a rough concave surface
vCis given by,
v_{c} = (R − r)ω
Also, v_{c} = rω_{1}
where ω_{1} is the rotational velocity of the sphere.
⇒ω1=vcr=Rrrω ⋯1As total energy of a particle in S.H.M. remains constant,
mgRr 1cos θ+12mvc2+12Iω12=constantSubstituting the values of vc and ω1 in the above equation, we get:mg Rr 1cos θ+12mRr2 ω2+12mr2Rrrω2=constant ∵ I = mr2mgRr 1cos θ+12mRr2 ω2+15mr2 Rrrω2=constant⇒gRr 1cos θ+Rr2ω2 12+15=constantTaking derivative on both sides, we get:
gRrsinθdθdt=710Rr22ωdωdt⇒gsinθ=2×710Rrα ∵ a=dωdt⇒gsinθ=75Rrα⇒α=5gsinθ7Rr =5gθ7Rr∴αθ=ω2=5g7Rr=constantTherefore, the motion is S.H.M.
ω=5g7RrTime period is given by,⇒T=2π7Rr5g
Question 40:
A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.
Answer:
It is given that:
Length of the pendulum, l = 40 cm = 0.4 m
Radius of the earth, R = 6400 km
Acceleration due to gravity on the earth’s surface, g = 9.8 ms
2
Let
g’be the acceleration due to gravity at a depth of 1600 km from the surface of the earth.
Its value is given by,
g’=g1dRwhere d is the depth from the earth surfce, R is the radius of earth, and g is acceleration due to gravity. On substituting the respective values, we get: g’=9.8116006400 =9.8114 =9.8×34=7.35 ms2
Time period is given as,
T=2πlg’
⇒T=2π0.47.35⇒T=2×3.14×0.23 =1.465≈1.47 s
Question 41:
Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of
gR(b) it is released from a height R above the tunnel (c) it is thrown vertically upward along the length of tunnel with a speed of
gR.
Answer:
Given:
Radius of the earth is R.
Let M be the total mass of the earth and
ρbe the density.
Let mass of the part of earth having radius x be M‘.
∴M’M=ρ×43πx3ρ×43πR3=x3R3⇒M’=Mx3R3Force on the particle is calculated as,
Fx=GM’mx2 =GMmR3x …1Now, acceleration
axof mass M‘ at that position is given by,
ax=GMR3x⇒axx=ω2=GMR3=gR ∵g=GMR2So, Time period of oscillation, T=2πRg(a) Velocitydisplacement equation in S.H.M is written as,
V=ωA2y2 where, A is the amplitude; and y is the displacement.When the particle is at y = R,
The velocity of the particle is
gRand
ω=gR.
On substituting these values in the velocitydisplacement equation, we get:
gR=gRA2R2 ⇒R2=A2R2⇒A=2RLet t_{1} and t_{2} be the time taken by the particle to reach the positions X and Y.
Now, phase of the particle at point X will be greater than
π2but less than
π.
Also, the phase of the particle on reaching Y will be greater than
πbut less than
3π2.
Displacementtime relation is given by,
y = A sin ωt
Substituting y = R and A =
2R, in the above relation, we get:
R=2R sin ωt1
⇒ωt1=3π4Also,
R=2R sin ωt2 ⇒ωt2=5π4So, ωt2t1=π2⇒t2t1=π2ω=π2gRTime taken by the particle to travel from X to Y:
t2t1=π2ω=π2Rgs
(b) When the body is dropped from a height R
Using the principle of conservation of energy, we get:
Change in P.E. = Gain in K.E.
⇒GMmRGMm2R=12mv2⇒v=gR As the velocity is same as that at X, the body will take the same time to travel XY.
(c) The body is projected vertically upwards from the point X with a velocity
gR. Its velocity becomes zero as it reaches the highest point.
The velocity of the body as it reaches X again will be,
v=gR Hence, the body will take same time i.e.
π2Rgs to travel XY.
Question 42:
Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R/2 from the earth’s centre where R is the radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational force exerted by the earth on a particle of mass m placed in the tunnel at a distance x from the centre of the tunnel. (b) Find the component of this force along the tunnel and perpendicular to the tunnel. (c) Find the normal force exerted by the wall on the particle. (d) Find the resultant force on the particle. (e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period.
Answer:
If
ρis the density of the earth, then mass of the earth
Mis given by,
M=43πR3ρSimilarly, mass M’ of the part of earth having radius x is given by,M’=43πx13ρM’=MR3x13(a) Let F be the gravitational force exerted by the earth on the particle of mass m. Then, its value is given by,
F=GM’mx12Substituting the value of M’ in the above equaation, we get:F=GMmR3x13x12 =GMmR3x1=GMmR3x2+R24(b)
Fy=Fcosθ =GMmx1R3xx1=GMmxR3Fx=Fsin θ =GMmx1R3R2x1=GMm2R2 (c)
Fx=GMmR2
∵Normal force exerted by the wall N = F_{x}
(d)The resultant force is
GMmxR3(e) Acceleration = Driving force/mass
=GMmxR3m=GMxR3
⇒a
∝ x (the body executes S.H.M.)
ax=ω2=GMR3⇒ω=GmR3⇒T=2πR3GM
Question 43:
A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with and acceleration a_{0} (b) is going down with an acceleration a_{0} and (c) is moving with a uniform velocity.
Answer:
The length of the simple pendulum is l.
â€‹Let x be the displacement of the simple pendulum..
(a)
From the diagram, the driving forces f is given by,
f = m(g + a_{0})sinθ …(1)
Acceleration (a) of the elevator is given by,
a=fm =g+a0sinθ =g+a0xl From the diagram sinθ=xl[ when θ is very small, sin θ → θ = x/l]
∴a=g+a0lx …(2)
As the acceleration is directly proportional to displacement, the pendulum executes S.H.M.
Comparing equation (2) with the expression a =
ω2x, we get:
ω2=g+a0lThus, time period of small oscillations when elevator is going upward(T) will be:
T=2πlg+a0(b)
When the elevator moves downwards with acceleration a_{0},
Driving force (F) is given by,
F = m(g − a_{0})sinθ
On comparing the above equation with the expression, F = ma,
Acceleration, a=ga0 sinθ=ga0xl=ω2xTime period of elevator when it is moving downwardT’ is given by,T’=2πω=2πlga0(c) When the elevator moves with uniform velocity, i.e. a_{0} = 0,
For a simple pendulum, the driving force
F is given by,
F=mgxlComparing the above equation with the expression, F=ma, we get:a=gxl⇒xa=lgT=2πdisplacementAcceleration =2πlg
Question 44:
A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes π/3 seconds to complete one oscillation. Find the acceleration of the elevator.
Answer:
It is given that:
Length of the simple pendulum, l = 1 feet
Time period of simple pendulum, T =
π3 sAcceleration due to gravity, g = 32 ft/s^{2}
^{â€‹}
^{â€‹}Let a be the acceleration of the elevator while moving upwards.
Driving force
f is given by,
f = m(g + a)sinθ
Comparing the above equation with the expression, f = ma, we get:
Acceleration, a = (g + a)sinθ = (g +a)θ (For small angle θ, sin θ → θ)
=g+axl=ω2x (From the diagram
θ=xl)
⇒ω= g+alTime period
Tis given as,
T=2πlg+aOn substituting the respective values in the above formula, we get:
π3=2π132+a19=4132+a⇒32+a=36⇒a=3632=4 ft/s2
Question 45:
A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.
Answer:
It is given that:
When the car is moving uniformly, time period of simple pendulum, T = 4.0 s
As the accelerator is pressed, new time period of the pendulum, T’ = 3.99 s
Time period of simple pendulum, when the car is moving uniformly on a horizontal road is given by,
T=2πlg⇒4=2πlgLet the acceleration of the car be a.
The time period of pendulum, when the car is accelerated, is given by:
T’=2πlg2+a212⇒3.99=2πlg2+a212Taking the ratio of T to T’, we get:TT’=43.99=g2+a21/4gOn solving the above equation for a, we get:
a=g10 ms2
Question 47:
The earring of a lady shown in figure (12−E18) has a 3 cm long light suspension wire. (a) Find the time period of small oscillations if the lady is standing on the ground. (b) The lady now sits in a merrygoround moving at 4 m s^{−1} in a circle of radius 2 m. Find the time period of small oscillations of the earring.
Figure
Answer:
Given,
Length of the long, light suspension wire, l = 3 cm = 0.03 m
â€‹Acceleration due to gravity, g = 9.8 ms
2
(a) Time period
Tis given by
T=2πlg =2π0.039.8 =0.34 second(b) Velocity of merrygoround, v = 4 ms
1
Radius of circle, r = 2 m
As the lady sits on the merrygoround, her earring experiences centripetal acceleration.
Centripetal acceleration
a is given by,
a=v2r=422=8 m/s2 Resultant acceleration
Ais given by,
A=g2+a2 =96.04+64 =12.65 m/s2 Time period,
T=2πlA
=2π0.0312.65 =0.30 second
Question 48:
Find the time period of small oscillations of the following systems. (a) A metre stick suspended through the 20 cm mark. (b) A ring of mass m and radius r suspended through a point on its periphery. (c) A uniform square plate of edge a suspended through a corner. (d) A uniform disc of mass m and radius r suspended through a point r/2 away from the centre.
Answer:
(a) Moment of inertia
Iabout the point X is given by,
I = I_{C.G} + mh^{2}
=ml212+mh2=ml212+m0.32=m112+0.09=m1+1.0812=m2.0812The time period
Tis given by,
T=2πImglwhere I= the moment of inertia, and l = distance between the centre of gravity and the point of suspension.On substituting the respective values in the above formula, we get:T=2π2.08 mm×12×9.8×0.3 =1.52 s(b) Moment of inertia
Iabout A is given as,
I = I_{C.G.} + mr^{2} = mr^{2} + mr^{2} = 2mr^{2}
The time period (T) will be,
T=2πImglOn substituting the respective values in the above equation, we have:T=2π2mr2mgr =2π2rg(c) Let I be the moment of inertia of a uniform square plate suspended through a corner.
I=ma2+a23=2m3a2
In the
△ABC, l^{2} + l^{2} = a^{2}
∴l=a2⇒T=2πImgl =2π2ma23mgl =2π2a2 3ga2 =2π8a3g
(d)
We know h=r2
Distance between the C.G. and suspension point, l=r2 Moment of inertia about A will be:
l = I_{C.G.} + mh^{2}
=mr22+mr22=mr212+14=34mr2Time period (T) will be,
T=2πImgl =2π3mr24mgl =2π3r2g
Question 49:
A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the road.
Answer:
It is given that the length of the rod is l.
Let point A be the suspension point and point B be the centre of gravity.
Separation between the point of suspension and the centre of mass, l’ =
l2Also, h =
l2Using parallel axis theorem, the moment of inertia about A is given as,
I=ICG+mh2 =ml212+ml24=ml23the time period T is given by,T=2πImgl’=2πImgl2 =2π2ml23mgl=2π2l3gLet T‘ be the time period of simple pendulum of length x.
Time period
(T’)is given by,
T’=2πxgAs the time period of the simple pendulum is equal to the time period of the rod,T’=T⇒2l3g=xg⇒x=2l3
Question 50:
A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period?
Answer:
Let m be the mass of the disc and r be its radius.
Consider a point at a distance x from the centre of gravity.
Thus, l = x
Moment of intertia
Iabout the point x will be,
I = I_{C.G} +mx^{2}
=mr22+mx2=mr22+x2Time period(T) is given as,
T=2πImglOn substituting the respective values in the above equation, we get:T=2πmr22+x2mgx (l = x) =2πmr2+2x22mgx =2πr2+2x22gx …(1)To determine the minimum value of T,
d2Tdx2=0
Now,d2Tdx2=ddx4π2r22gx+4π22x22gx⇒2π2r2g1×2+4π2g=0⇒π2r2gx2+2π2g=0⇒π2r2gx2=2π2g⇒2×2=r2⇒x=r2
Substituting this value of x in equation (1), we get:
T=2πr2+2r222gx =2π2r22gx=2πr2gr2 =2π2r2gr==2π2rg
Question 51:
A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?
Answer:
It is given that:
Radius of the hollow sphere, r = 2 cm
Length of the long thread, l = 18 cm =
18100=0.18 m=0.2 m.
Let I be the moment of inertia and
ωbe the angular speed.
Using the energy equation, we can write:
mgl(1cos θ)+12Iω2=constant
mg0.20 1cos θ+12Iω2=C …1Moment of inertia about the point of suspension A is given by,I=23mr2+ml2Substituting the value of l in the above equation, we get:I=23m0.022+m 0.22 =23m0.0004+m0.04 =m0.00083+0.04 =m0.12083`
On substituting the value of I in equation (1) and differentiating it, we get:
ddtmg 0.2 1cos θ+120.12083mω2=ddtc⇒mg0.2sinθdθdt+120.12083m×2ωdωdt=0 ⇒2sin θ=0.12083α because, g=10 m/s2⇒αθ=60.1208⇒ω2=49.66⇒ω=7.04Thus, time periodT will be:T=2πω=0.89 sFor a simple pendulum, time period (T) is given by,
T=2πlg
⇒T= 2π0.1810= 0.86 s
% change in the value of time period=0.890.860.89×100=0.3∴ It is about 0.3% greater than the calculated value.
Question 52:
A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2^{0} and time period 2 s. Find (a) the radius of the circular wire, (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position, (c) the acceleration of this particle as it goes through its mean position and (d) the acceleration of this particle when it is at an extreme position. Take g = π^{2} m s^{−2}.
Answer:
It is given that:
Time period of oscillation, T = 2 s
Acceleration due to gravity, g =
π2ms^{−2}
Let I be the moment of inertia of the circular wire having mass m and radius r.
(a) Time period of compound pendulum
Tis given by,
T=2Imgl=2Imgr
∵l=r …(1)
Moment of inertia about the point of suspension is calculated as,
I = mr^{2} + mr^{2} = 2mr^{2}
On substituting the value of moment of inertia I in equation (1), we get:
T=2π2mr2mgr=2π2rg⇒22π=2rg⇒2rg=1π2⇒r=g2π2 =0.5 m=50 cm(b) From the energy equation, we have:
12Iω20=mgr 1cos θ12Iω20=mgr 1cos 2° ⇒122mr2·ω2=mgr1cos 2° ∵I=2mr2⇒ω2=gr1cos 2°On substituing the value of g and r in the above equation, we get:ω=0.11 rad/s ⇒v=ω×2r=11 cms1
(c) The acceleration is found to be centripetal at the extreme position.
Centripetal acceleration at the extreme position
anis given by,
a_{n} = ω^{2}(2r) = (0.11) × 100 = 12 cm/s^{2}
The direction of a_{n} is towards the point of suspension.
(d) The particle has zero centripetal acceleration at the extreme position. However, the particle will still have acceleration due to the S.H.M.
Angular frequency
ωis given by,
ω=2πT =2π2=3.14
∴Angular acceleration
aat the extreme position is given as,
α=ω2θα=ω22°=π2×2π180 =2π3180 1°=π180radianThus, tangential acceleration
=α2r=2π3180×100 = 34 cm/s^{2}
Question 53:
A uniform disc of mass m and radius r is suspended through a wire attached to its centre. If the time period of the torsional oscillations be T, what is the torsional constant of the wire?
Answer:
It is given that:
Mass of disc = m
Radius of disc = r
The time period of torsional oscillations is T.
Moment of inertia of the disc at the centre, I
=mr22Time period of torsional pendulum
Tis given by,
T=2πIkwhere I is the moment of inertia, and
k is the torsional constant.
On substituting the value of moment of inertia in the expression for time period T, we have:
T=2πmr22k On squaring both the sides, we get:T2=4π2mr22k=2π2mr2k⇒2π2mr2=kT2⇒k=2π2mr2T2
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Question 54:
Two small balls, each of mass m are connected by a light rigid rod of length L. The system is suspended from its centre by a thin wire of torsional constant k. The rod is rotated about the wire through an angle θ_{0} and released. Find the force exerted by the rod on one of the balls as the system passes through the mean position.
Figure
Answer:
It is given that the mass of both the balls is m and they are connected to each other with the help of a light rod of length L.
Moment of inertia of the twoball system
Iis given by,
I=2mL22=mL22
Torque
τ, produced at any given position θ is given as:
τ= kθ
⇒Work done during the displacement of system from 0 to θ_{0} will be,
W=∫0θ0kθ dθ=kθ022On applying workenergy theorem, we get:
12Iω20=work done=kθ022∴ω2=kθ02I=kθ02mL2
From the free body diagram of the rod, we can write:
Force, T2=mω2L2+mg2 =mkθ02mL2×L2+m2g2 =k2θ04L2+m2g2
Question 55:
A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0°, (b) 60°, (c) 90°.
Answer:
It is given that a particle is subjected to two S.H.M.s of same time period in the same direction.
Amplitude of first motion, A_{1} = 3 cm
Amplitude of second motion, A_{2} = 4 cm
Let Ï• be the phase difference.
The resultant amplitude
Ris given by,
R=A12+A22+2A1A2 cos ϕ(a) When Ï• = 0°
R=32+42+234 cos 0° =7 cm(b) When Ï• = 60°
R=32+42+234 cos 60° =37=6.1 cm(c) When Ï• = 90°
R=32+42+234cos 90° =25=5 cm
Question 56:
Three simple harmonic motions of equal amplitude A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion.
Answer:
It is given that three S.H.M.s of equal amplitudes A and equal time periods are combined in the same direction.
Let
Y1, Y2 and Y3be the three vectors representing the motions, as shown in the figure given below.
According to the question:
Angle betweenY1 and Y2 = 60 °Angle between Y2 and Y3 = 60 °.
By using the vector method, we can find the resultant vector.
Resultant amplitude = Vector sum of the three vectors
= A + A cos 60° + A cos 60°
=A+A2+A2=2A
Question 57:
A particle is subjected to two simple harmonic motions given by x_{1} = 2.0 sin (100π t) and x_{2} = 2.0 sin (120 π t + π/3), where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025.
Answer:
Given are the equations of motion of a particle:
x_{1} = 2.0sin100
πt
x2=2.0sin120πt+π3The resultant displacement
xwill be,
x = x_{1} + x_{2}
=2sin100πt+sin120πt+π3(a) At t = 0.0125 s
x=2sin100π ×0.0125+sin120π ×0.0125+π3 =2sin 5π4+sin 3π2+π3 =20.707+0.5 =2×1.207=2.41 cm(b) At t = 0.025 s
x=2sin100 π×0.025+sin120π×0.025+π3 =2sin10π4+sin3π+π3 =21+0.866 =2×0.134=0.27 cm
Question 58:
A particle is subjected to two simple harmonic motions, one along the Xaxis and the other on a line making an angle of 45° with the Xaxis. The two motions are given by x = x_{0} sin ωt and s = s_{0} sin ωt. Find the amplitude of the resultant motion.
Answer:
Given:
Equation of motion along X axis, x = x_{0}sinωt
Equation of motion along Y axis, s = s_{0}sinωt
Angle between the two motions,
θ= 45^{â‚€}
Resultant motion (R) will be,
R=x2+s2+2xscos45° =x02sinωt+s02sinωt +2x0s0sin2ωt12 =x02+s02+2x0s01/2sinωtHence, the resultant amplitude is
x02+s02+2x0s01/2.
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity
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