HC Verma Solutions for Class 11 Physics Chapter 15 – Wave Motion and Wave on a String

Page No 321:

Question 1:

You are walking along a seashore and a mild wind is blowing. Is the motion of air a wave motion?

Answer:

No, in wave motion there is no actual transfer of matter but transfer of energy between the points where as when wind blows air particles moves with it.

Question 2:

The radio and TV programmes, telecast at the studio, reach our antenna by wave motion. Is it a mechanical wave or nonmechanical?

Answer:

It is a non-mechanical wave because this type of wave does not require a material medium to travel.

Question 3:

A wave is represented by an equation

y=c1 sin c2x+c3t. In which direction is the wave going? Assume that

c1, c2and

c3are all positive.

Answer:

Equation of the wave is

y=c1 sin c2x+c3tWhen the variable of the equation is (c2x + c3t), then the wave must be moving in the negative x-axis with time t.

Question 4:

Show that the particle speed can never be equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by 2π.

Answer:

Equation of the wave is given by

y=Asinωt-kxwhere
A is the amplitude
ω is the angular frequency
k is the wave number
Velocity of wave,

v=ωkVelocity of particle,

vp=dydt=Aω cosωt-kxMax velocity of particle,

vpmax=AωAs given

A<λ2π

vpmax=λω2πvpmax<ωk             ∵2πλ=k

Question 5:

Two wave pulses identical in shape but inverted with respect to each other are produced at the two ends of a stretched string. At an instant when the pulses reach the middle, the string becomes completely straight. What happens to the energy of the two pulses?

Answer:

When two wave pulses identical in shape but inverted with respect to each other meet at any instant, they form a destructive interference. The complete energy of the system at that instant is stored in the form of potential energy within it. After passing each other, both the pulses regain their original shape.

Question 6:

Show that for a wave travelling on a string

ymaxνmax=νmaxαmax,where the symbols have usual meanings. Can we use componendo and dividendo taught in algebra to write

ymax+νmaxνmax-νmax=νmax+αmaxνmax-αmax?

Answer:

ymaxvmax=vmaxamaxy=Asinωt-kxv=dydt=Acosωt-kxvmax=Aωa=dvdt=-Aω2sinωt-kxamax=ω2ATo prove,

ymaxvmax=vmaxamaxLHSymaxvmax=AAω=1ωRHSvmaxamax=Aωω2A=1ωNo, componendo and dividendo is not applicable. We cannot add quantities of different dimensions.

Question 7:

What is the smallest positive phase constant which is equivalent to 7⋅5 π?

Answer:

Equation of the wave: y = sin(kxωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.

The argument of the sine is a phase, so the smallest positive phase constant should be

sin7.5π=sin3×2π+1.5π                =sin1.5πTherefore, the smallest positive phase constant is 1.5π.

Question 8:

A string clamped at both ends vibrates in its fundamental mode. Is there any position (except the ends) on the string which can be touched without disturbing the motion? What if the string vibrates in its first overtone?

Answer:

Yes, at the centre. The centre position is a node. If the string vibrates in its first overtone, then there will be two positions, i.e., two nodes, one at x = 0 and the other at x = L.

Page No 322:

Question 1:

A sine wave is travelling in a medium. The minimum distance between the two particles, always having same speed, is
(a)

λ/4(b)

λ/3(c)

λ/2(d)

λ.

Answer:

(c)

λ/2
A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the positions of maxima and minima is equal to

λ/2.

Question 2:

A sine wave is travelling in a medium. A particular particle has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distance
(a)

λ/4(b)

λ/3(c)

λ/2(d)

λ.

Answer:

(c)

λ/2A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant, then the particle closest to it having zero displacement is at a distance is equal to

λ/2.

Question 3:

Which of the following equations represents a wave travelling along Y-axis?
(a)

x=A sin ky-ωt(b)

y=A sin kx-ωt(c)

y=A sin ky cos ωt(d)

y=A cos ky sin ωt.

Answer:

(a)

x=A sin ky-ωtHere x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.

Question 4:

The equation

y=A sin2 kx-ωtrepresents a wave motion with
(a) amplitude A, frequency

ω/2π(b) amplitude A/2, frequency

ω/π(c) amplitude 2A, frequency

ω/4π(d) does not represent a wave motion.

Answer:

(b) amplitude A/2, frequency

ω/π y=Asin2kx-ωt

cos2θ=1-2sin2θsin2θ=1-cos2θ2 y=A1-cos2kx-ωt2y=A21-cos2kx-ωtThus, we have:
Amplitude =

A2Frequency =

2ω2π=ωπ

Question 5:

Which of the following is a mechanical wave?
(a) Radio waves
(b) X-rays
(c) Light waves
(d) Sound waves.

Answer:

(d) Sound waves

There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound requires medium to travel, hence it is a mechanical wave.

Question 6:

A cork floating in a calm pond executes simple harmonic motion of frequency

νwhen a wave generated by a boat passes by it. The frequency of the wave is
(a)

ν(b)

ν/2(c)

2ν(d)

2ν.

Answer:

(a)

νThe boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.

Question 7:

Two strings A and B, made of same material, are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed

νAand on B with speed

νB. The ratio

νA/νBis
(a) 1/2
(b) 2
(c) 1/4
(d) 4.

Answer:

(a) 1/2

Wave speed is given by

ν=Tµwhere
​T is the tension in the string
v is the speed of the wave
μ is the mass per unit length of the string

µ=ML=ρVL=ρALLwhere
M is the mass of the string, which can be written as ρV
L is the length of the string

=ρπr2=ρπD24∴ν=TρπD24=2DTρπwhere D is the diameter of the string.
Thus, v

1DSince, rA = 2rB

vA∝12rA∝12×2rB                      (1)vB∝12rB                                      (2)From Equations (1) and (2) we get

vAvB=12

Question 8:

Both the strings, shown in figure (15-Q1), are made of same material and have same cross section. The pulleys are light. The wave speed of a transverse wave in the string AB is

ν1and in CD it is

ν2. Then

ν1/ν2is
(a) 1
(b) 2
(c)

2(d)

1/2.
Figure

Answer:

(d)

1/2

TAB=TTCD=2Twhere
​TAB is the tension in the string AB
​TCD is the tension in the string CD
The eelation between tension and the wave speed is given by

v=Tµv∝Twhere
v is the wave speed of the transverse wave
μ is the mass per unit length of the string

v1v2=T2T=12

Question 9:

Velocity of sound in air is 332 m s−1. Its velocity in vacuum will be
(a) > 332 m s1
(b) = 332 m s−1
(c) < 332 m s−1
(d) meaningless.

Answer:

(d) meaningless

Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.

Question 10:

A wave pulse, travelling on a two-piece string, gets partially reflected and partially transmitted at the junction. The reflected wave is inverted in shape as compared to the incident one. If the incident wave has wavelength

λand the transmitted wave

λ’.
(a)

λ’>λ(b)

λ’=λ(c)

λ'<λ(d) nothing can be said about the relation of

λ and λ’.

Answer:

(c)

λ'<λ
As

v=fµA wave pulse travels faster in a thinner string.
The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.

Question 11:

Two waves represented by

y=asinωt-kxand

y=acosωt-kxare superposed. The resultant wave will have an amplitude
(a) a
(b)

2a(c) 2a
(d) 0.

Answer:

(b)

2aWe know that the resultant of the amplitude is given by

Rnet=A12+A22+2A1A2 cosϕFor the particular case, we can write

=a2+a2+2a2cosπ2=2a

Question 12:

Two wires A and B, having identical geometrical construction, are stretched from their natural length by small but equal amount. The Young modules of the wires are YA and YB whereas the densities are

ρA and ρB. It is given that YA > YB and

ρA>ρB. A transverse signal started at one end takes a time t1 to reach the other end for A and t2 for B.
(a) t1 < t2
(b) t1 = t2
(c) t1 > t2
(d) the information is insufficient to find the relation between t1 and t2.

Answer:

(d) the information is insufficient to find the relation between t1 and t2.

v=ηρBut because the length of wires A and B is not known, the relation between A and B cannot be determined.

Question 13:

Consider two waves passing through the same string. Principle of superposition for displacement says that the net displacement of a particle on the string is sum of the displacements produced by the two waves individually. Suppose we state similar principles for the net velocity of the particle and the net kinetic energy of the particle. Such a principle will be valid for
(a) both the velocity and the kinetic energy
(b) the velocity but not for the kinetic energy
(c) the kinetic energy but not for the velocity
(d) neither the velocity nor the kinetic energy.

Answer:

(b) the velocity but not for the kinetic energy

The principle of superposition is valid only for vector quantities. Velocity is a vector quantity, but kinetic energy is a scalar quantity.

Question 14:

Two wave pulses travel in opposite directions on a string and approach each other. The shape of one pulse is inverted with respect to the other.
(a) The pulses will collide with each other and vanish after collision.
(b) The pulses will reflect from each other, i.e., the pulse going towards right will finally move towards left and vice versa.
(c) The pulses will pass through each other but their shapes will be modified.
(d) The pulses will pass through each other without any change in their shapes.

Answer:

(d) The pulses will pass through each other without any change in their shapes.

The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.

Question 15:

Two periodic waves of amplitudes A1 and A2 pass thorough a region. If A1 > A2, the difference in the maximum and minimum resultant amplitude possible is
(a) 2A1
(b) 2A2
(c) A1 + A2
(d) A1A2.

Answer:

(b) 2A2

We know resultant amplitude is given by

Anet=A12+A22+2A1A2cosϕFor maximum resultant amplitude

Amax=A1+A2For minimum resultant amplitude

Amin=A1-A2So, the difference between Amax and Amin is

Amax-Amin=A1+A2-A1+A2=2A2

Question 16:

Two waves of equal amplitude A, and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is
(a) 0
(b) A
(c) 2A
(d) between 0 and 2A.

Answer:

(d) between 0 and 2A

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle (φ). So, the resultant amplitude lies between the maximum resultant amplitude (Amax) and the minimum resultant amplitude (Amin).
Amax = A + A = 2A
Amin = AA = 0

Question 17:

Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120°. The resultant amplitude will be
(a) A
(b) 2A
(c) 4A
(d)

2A.

Answer:

(a) A

We know the resultant amplitude is given by

Rnet=A2+A2+2A2 cos 120º  (ϕ=120°)=2A2-A2              ∵ cos 120º=-12=A

Question 18:

The fundamental frequency of a string is proportional to
(a) inverse of its length
(b) the diameter
(c) the tension
(d) the density.

Answer:

(a) inverse of its length

The relation between wave speed and the length of the string is given by

v=12lFµwhere
l is the length of the string
F is the tension
μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.

v∝1l

Page No 323:

Question 1:

A wave pulse passing on a string with a speed of 40 cm s−1 in the negative x-direction has its maximum at x = 0 at t = 0. Where will this maximum be located at t = 5 s?

Answer:


Given,
Speed of the wave pulse passing on a string in the negative x-direction = 40 cms−1
As the speed of the wave is constant, the location of the maximum after 5 s will be
s = v × t
= 40 × 5
= 200 cm (along the negative x-axis)
Therefore, the required maximum will be located after x = −2 m.

Question 2:

The equation of a wave travelling on a string stretched along the X-axis is given by

y=A e -xa+tT2.(a) Write the dimensions of A, a and T. (b) Find the wave speed. (c) In which direction is the wave travelling? (d) Where is the maximum of the pulse located at t = T? At t = 2 T?

Answer:

Given,
Equation of the wave travelling on a string stretched along the X-axis:

y=Ae xa+tT-2(a) The dimensions of A (amplitude), T (time period) and

a=λ2π, which will have the dimensions of the wavelength, are as follows:

A = M0L1T0T=M0L0T-1a=M0L1T0(b) Wave speed,

ν=λT=aT        λ=a(c) If

y=f t+xν, then the wave travels in the negative direction; and if

y=f t-xν, then the wave travels in the positive direction.
Thus, we have:

y=Ae xa+tT-2   =Ae-1Tt+xTa2   =Ae-1Tt+xV   = Ae-ft+xVHence, the wave is travelling is the negative direction.

(d) Wave speed,

v=atMaximum pulse at t = T  =

aT×T=a        Along the negative x-axisMaximum pulse at t = 2T =

aT×2T=2a       Along the negative x-axisTherefore, the wave is travelling in the negative x-direction.

Question 3:

Figure (15-E1) shows a wave pulse at t = 0. The pulse moves to the right with a speed of 10 cm s−1. Sketch the shape of the string at t = 1 s, 2 s and 3 s.

Figure

Answer:

Given,
Wave pulse at t = 0

Wave speed = 10 cms−1
Using the formula

s=v×t, we get:

At:t=1 s, s1=ν×t=10×1=10 cmt = 2 s, s2=ν×t=10×2=20 cmt = 3 s, s3=ν×t=10×3=30 cm

Question 19:

A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 240 Hz. The wire will vibrate with a frequency of
(a) 240 Hz
(b) 480 Hz
(c) 720 Hz
(d) will not vibrate.

Answer:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.

Question 20:

A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 410 Hz. The wire will vibrate with a frequency
(a) 410 Hz
(b) 480 Hz
(c) 820 Hz
(d) 960 Hz.

Answer:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.

Question 21:

A sonometer wire of length l vibrates in fundamental mode when excited by a tuning fork of frequency 416. Hz. If the length is doubled keeping other things same, the string will
(a) vibrate with a frequency of 416 Hz
(b) vibrate with a frequency of 208 Hz
(c) vibrate with a frequency of 832 Hz
(d) stop vibrating.

Answer:

(b) vibrate with a frequency of 208 Hz

According to the relation of the fundamental frequency of a string

ν=12lFμwhere
l is the length of the string
F is the tension
μ is the linear mass density

We know that ν1 = 416 Hz, l1 = l and l2 = 2l.

v1∝1l1v1l1=v2l2416l=v22lv2=208 Hz

Question 22:

A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tuning fork of frequency 416. Hz. The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to
(a) 1 kg
(b) 2 kg
(c) 8 kg
(d) 16 kg.

Answer:

(d) 16 kg

According to the relation of the fundamental frequency of a string

ν=12lFμwhere l is the length of the string
F is the tension
μ is the linear mass density of the string
We know that ν1 = 416 Hz, l1 = l and l2 = 2l.
Also, m1 = 4 kg and m2 = ?

ν1=12l1m1gμ          (1)

ν2=12l2m2gμ        (2)So, in order to maintain the same fundamental mode

ν1=ν2squaring both sides of equations (1) and (2) and then equating

14l24gμ = 116l2m2gμ⇒m2=16 kg

Question 1:

A mechanical wave propagates in a medium along the X-axis. The particles of the medium
(a) must move on the X-axis
(b) must move on the Y-axis
(c) may move on the X-axis
(d) may move on the Y-axis.

Answer:

(c) may move on the X-axis
(d) may move on the Y-axis

A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.

Question 2:

A transverse wave travels along the Z-axis. The particles of the medium must move
(a) along the Z-axis
(b) along the X-axis
(c) along the Y-axis
(d) in the XY plane.

Answer:

(d) in the XY plane

In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the XY plane.

Question 3:

Longitudinal waves cannot
(a) have a unique wavelength
(b) transmit energy
(c) have a unique wave velocity
(d) be polarized.

Answer:

(d) be polarised

A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.

Question 4:

A wave going in a solid
(a) must be longitudinal
(b) may be longitudinal
(c) must be transverse
(d) may be transverse.

Answer:

(b) may be longitudinal
(d) may be transverse

Particles in a solid are very close to each other; thus, both longitudinal and transverse waves can travel through it.

Question 5:

A wave moving in a gas
(a) must be longitudinal
(b) may be longitudinal
(c) must transverse
(d) may be transverse.

Answer:

(a) must be longitudinal

Because particles in a gas are far apart, only longitudinal wave can travel through it.

Question 6:

Two particles A and B have a phase difference of π when a sine wave passes through the region.
(a) A oscillates at half the frequency of B.
(b) A and B move in opposite directions.
(c) A and B must be separated by half of the wavelength.
(d) The displacements at A and B have equal magnitudes.

Answer:

(b) A and B move in opposite directions.
(d) The displacements at A and B have equal magnitudes.


A and B have a phase difference of π. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.

yA→=Asinωt

yB→=Bsinωt+π

Question 7:

A wave is represented by the equation
y=0·001 mm sin50 s-1t+2·0 m-1x.

(a) The wave velocity = 100 m s−1.
(b) The wavelength = 2⋅0 m.
(c) The frequency = 25/π Hz.
(d) The amplitude = 0⋅001 mm.

Answer:

(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm

y=0·001 mm sin50 s-1t+2·0 m-1xEquating the above equation with the general equation, we get:

y=Asinωt-kxω=2πT=2πvk=2πλHere, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.

A=0.001 mmNow,50=2πν⇒ν=25π Hz

Question 8:

A standing wave is produced on a string clamped at one end and free at the other. The length of the string
(a) must be an integral multiple of

λ/4(b) must be an integral multiple of

λ/2(c) must be an integral multiple of

λ(d) must be an integral multiple of

λ/2.

Answer:

(a) must be an integral multiple of

λ/4A standing wave is produced on a string clamped at one end and free at the other. Its fundamental frequency is given by

ν=n+12v2L⇒v=νλ⇒ν=n+12νλ2L⇒L=2n+14λ⇒L=λ4,3λ4,…

Question 9:

Mark out the correct options.
(a) The energy of any small part of a string remains constant in a travelling wave.
(b) The energy of any small part of a string remains constant in a standing wave.
(c) The energies of all the small parts of equal length are equal in a travelling wave.
(d) The energies of all the small parts of equal length are equal in a standing wave.

Answer:

(b) The energy of any small part of a string remains constant in a standing wave.

A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.

Question 10:

In a stationary wave,
(a) all the particles of the medium vibrate in phase
(b) all the antinodes vibrates in phase
(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase.

Answer:

(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase

All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.

Thus, particles in alternate antinodes vibrate in the same phase.

Page No 324:

Question 4:

A pulse travelling on a string is represented by the function

y=a2x-νt2+a2,where a = 5 mm and

ν=20 cm-1. Sketch the shape of the string at t = 0, 1 s and 2 s. Take x = 0 in the middle of the string.

Answer:

Given,
Pulse travelling on a string,

y=a3x-νt2+a2

a=5 mm=0.5 cmWave speed, ν = 20 cm/sSo, at

t=0 s, y=a3x2+a2.
Similarly, at t = 1 s,

y=a3x-ν2+a2And, At t=2 s,y=a3x-2ν2+a2To sketch the shape of the string, we have to plot a graph between y and x at different values of t.

Question 5:

The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive x-direction is given f(t) = A sin (t/T). The wave speed is

ν. Write the wave equation.

Answer:

Given,
Equation of the wave travelling in the positive x-direction at x = 0:

ft=AsintTHere,
Wave speed = v
Wavelength, λ = vT
T = Time period
Therefore, the general equation of the wave can be represented by

y=AsintT-xνT

Question 6:

A wave pulse is travelling on a string with a speed

νtowards the positive X-axis. The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
(a) What are the dimensions of A and a ? (b) Write the equation of the wave for a general time t, if the wave speed is

ν.

Answer:

The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
Dimensions of A and a are governed by the dimensional homogeneity of the equation g(x) = A sin(x/a).
Now,

(a)  M0L1T0=A⇒A=LAnd, a=M0L1T0⇒a=L(b) Wave speed=ν∴ Time period, T=aνHere, a=Wave length=λ The general equation of wave is represented byy=Asinxa-tav   =Asinx-νta

Question 7:

A wave propagates on a string in the positive x-direction at a velocity

ν. The shape of the string at

t=t0is given by

gx, t0=A sin x/a. Write the wave equation for a general time t.

Answer:

Given,
Wave velocity =

νShape of the string at

t=t0=

gx, t0=A sin x/a    …(i)
For a wave travelling in the positive x-direction, the general equation is given by

y=A sin xa-tTPutting t = − t and comparing with equation (i), we get:

gx,0=Asinxa+t0T⇒gx,t=Asinxa+t0T-tTNow,T=aνHere, a=Wave lengthν=Velocity of the waveThus, we have:y=Asin xa+t0aν-taν⇒y=Asin x+ν t0-ta

Question 8:

The equation of a wave travelling on a string is

y=0·10 mm sin31·4 m-1x+314 s-1t.

(a) In which direction does the wave travel? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string?

Answer:

Given,
Equation of the wave,

y=0.10 mm sin31.4 m-1x+314 s-1 tThe general equation is

y=Asin2πxλ+ωt.
From the above equation, we can conclude:

(a) The wave is travelling in the negative x-direction.

(b)

2πλ=31.4 m-1

⇒λ=2π31.4=0.2 m= 20 cmAnd, ω=314 s-1⇒2πf=314⇒f=3142π      =3142×3.14      =50 s-1=50 HzWave speed:

ν=λf=20×50  = 1000 cm/s(c) Maximum displacement, A = 0.10 mm

Maximum velocity=aω=0.1×10-1×314                                  =3.14 cm/s

Question 9:

A wave travels along the positive x-direction with a speed of 20 m s−1. The amplitude of the wave is 0⋅20 cm and the wavelength 2⋅0 cm. (a) Write the suitable wave equation which describes this wave. (b) What is the displacement and velocity of the particle at x = 2⋅0 cm at time t = 0 according to the wave equation written? Can you get different values of this quantity if the wave equation is written in a different fashion?

Answer:

A wave travels along the positive x-direction.
Wave amplitude (A) = 0.20 cm
Wavelength (λ) = 20 cm
Wave speed (v) = 20 m/s

(a) General wave equation along the x-axis:

y=Asinkx-ωt∴k=2πλ=2π2=π cm-1T=λν=22000    =11000=10-3 sω=2πT=2π×103 s-1Wave equation:

y=0.2 cm sinπ cm-1 x-2π×10-3 s-1(b) As per the question
For the wave equation ,we need to find the displacement and velocity at x = 2 cm and t = 0.

y=0.2 cm sin2π=0∴ν=Aωcosπx      =0.2×2000π×cos2π      =400π      =400π cm/s=4π m/sIf the wave equation is written in a different fashion, then also we will get the same values for these quantities.

Question 10:

A wave is described by the equation

y=1·0 mm sin πx2·0 cm-t0·01 s.(a) Find the time period and the wavelength? (b) Write the equation for the velocity of the particles. Find the speed of the particle at x = 1⋅0 cm at time t = 0⋅01 s. (c) What are the speeds of the particles at x = 3⋅0 cm, 5⋅0 cm and 7⋅0 cm at t = 0⋅01 s?
(d) What are the speeds of the particles at x = 1⋅0 cm at t = 0⋅011, 0⋅012, and 0⋅013 s?

Answer:

The wave equation is represented by

y=1·0 mm sin πx2·0 cm-t0·01 sLet:
Time period = T
Wavelength = λ

a T=2×0.01=0.02 s=20 ms      λ=2×2=4 cm(b) Equation for the velocity of the particle:

ν=dydt=ddtsin 2πx4-t0.02

=-0.50 cos 2πx4-t0.02×10.02⇒ν=-0.50 cos 2π x4-t0.02At x=1 and t=0.01 s, ν=-0.50 cos 2π 14-12=0.(c) (i) Speed of the particle:
At x=3 cm and t=0.01 s, ν=-0.50cos2π34-12=0.
(ii)

At x=5 cm and t=0.01 s,

ν=0 iii At  x=7 cm and t=0.1 s, ν=0.iv At x=1 cm and t=0.011 s,ν=50 cos 2π14-0.0110.02  =-50 cos 3π5=-9.7 cm/s(By changing the value of t, the other two can be calculated.)

Question 11:

A particle on a stretched string supporting a travelling wave, takes 5⋅0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2⋅0 cm. Find the frequency, the wavelength and the wave speed.

Answer:

Time taken to reach from the mean position to the extreme position,

T4= 5 ms
Time period (T) of the wave:

T=4×5 ms  =20×10-3=2×10-2 sWavelength (λ) =

2×Distance between two mean positions

=2×2 cm=4 cm

Frequency, f=1T                    =12×10-2                    =50 HzWave speed, v= λf                      =4×10-2×50                      =200×10-2                      =2 m/s

Question 12:

Figure (15-E2) shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is 20 cm s−1. Find (a) the amplitude, (b) the wavelength, (c) the wave number and (d) the frequency of the wave.

Figure

Answer:

Given:
Wave speed,

ν=20 cm/sFrom the graph, we can infer:
(a) Amplitude, A = 1 mm
(b) Wavelength, λ = 4 cm

(c) Wave number,

k=2πλ

=2×3.144=1.57 cm-2(d)

Time period, T=λν

Frequency, f=1T=vλ⇒f=204=5 Hz

Question 13:

A wave travelling on a string at a speed of 10 m s−1 causes each particle of the string to oscillate with a time period of 20 ms. (a) What is the wavelength of the wave? (b) If the displacement of a particle of 1⋅5 mm at a certain instant, what will be the displacement of a particle 10 cm away from it at the same instant?

Answer:

Given,
Wave speed (v) = 10 ms−1
Time period (T) = 20 ms

=20×10-3=2×10-2 s(a) Wavelength of the wave:

λ=νt=10×2×10-2  = 0.02 m=20 cm(b) Displacement of the particle at a certain instant:

y=asinωt-kx⇒1.5=asinωt-kxPhase difference of the particle at a distance x = 10 cm:

ϕ=2πxλ=2π×1020=π

The displacement is given by y’=asinωt-kx+π    =asinωt-kx=1.5 mm∴Displacement=1.5 mm

Question 14:

A steel wire of length 64 cm weighs 5 g. If it is stretched by a force of 8 N, what would be the speed of a transverse wave passing on it?

Answer:

Given,
Length of the steel wire = 64 cm
Weight = 5 g
Applied force = 8 N
Thus, we have:

Mass per unit length=564 gm/cmTension, T=8 N                   =8×105 dynSpeed, v=Tm               =8×105×645               =3200 cm/s=32 m/s

Question 15:

A string of length 20 cm and linear mass density 0⋅40 g cm−1 is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t = 0 near an ends as shown in figure (15-E3), which travels towards the other end. (a) When will the string have the shape shown in the figure again? (b) Sketch the shape of the string at a time half of that found in part (a).

Figure

Answer:


Given,
Length of the string = 20 cm
Linear mass density of the string = 0.40 g cm−1
Applied tension = 16 N =

16×105 dynVelocity of the wave:

ν=Tm  =16×1050.4  =2000 cm/s∴ Time taken to reach the other end

=202000=0.01 sTime taken to see the pulse again in the original position

=0.01×2=0.02 s(b) At t = 0.01 s, there will be a trough at the right end as it is reflected.

Question 16:

A string of linear mass density 0⋅5 g cm−1 and a total length 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end (figure 15-E4). The ring can move on a vertical rod. A wave pulse is produced on the string which moves towards the ring at a speed of 20 cm s−1. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring. (a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to region its shape. (b) The shape of the string changes periodically with time. Find this time period. (c) What is the tension in the string?

Figure

Answer:

Given,
Linear mass density of the string = 0.5 gcm−1
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms−1

The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.

Phase change=0(a)

Total distance, S=20+20=40 cmWave speed, ν=20 m/sTime taken to regain shape:

Time=Sν=4020=2 s(b) The wave regain its shape after covering a period distance

=2×30=60cm

∴ Time period=6020=3 s(c) Frequency,

n=1Time period=13 s-1We know:

n=12lTmHere, T is the tension in the string.
Now,

m=Mass per unit length    =0.5 gm/cm⇒13=12×30 T0.5⇒ T=400×0.5       =200 dyn       =2×10-3 N

Question 17:

Two wires of different densities but same area of cross section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.

Answer:

Let:
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v1.
Thus, we have:
​

ν1=Tm1The mass per unit length can be given as

m1=ρ1a1I1I1=ρ1a1⇒ν1=Tρ1a1      …(1)
Let the velocity of the first string be v 2.  
Thus, we have:

ν2=Tm2⇒ ν2=Tρ2a2     …(2)
Given,

ν1=2ν2⇒Ta1ρ1=2Ta2ρ2⇒Ta1ρ1=4 Ta2ρ2⇒ρ1ρ2=14⇒ρ1:ρ2 = 1:4

Page No 325:

Question 18:

A transverse wave described by

y=0·02 m sin 1·0 m-1 x+30 s-1tpropagates on a stretched string having a linear mass density of

1·2×10-4 kg m-1. Find

Answer:

Given,
Wave equation,

y=0·02 msin1·0 m-1x+30 s-1t Let: Mass per unit length, m=1.2×10-4 kg/mFrom the wave equation, we have:k=1 m-1=2πλAnd,ω=30 s-1=2πfVelocity of the wave in the stretched string is given by

ν=λf=ωk=301⇒v=30 m/sWe know:v=Tm⇒30 =T1.2×10-4 ⇒T=108×10-3=0.108 NSo, the tension in the string is 0.108 N.

Question 19:

A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1⋅0 and the displacement becomes zero 200 times per second. The linear mass density of the string is 0⋅10 kg m−1 and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10 ms.

Answer:

Given,
Amplitude of the wave = 1 cm
Frequency of the wave,

f=2002=100 HzMass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N

(a) Velocity of the wave is given by

v=TmThus, we have:

v=900.1=30 m/sNow,Wavelength, λ=vf=30100=0.3 m⇒λ=30 cm(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by

y=1 cmcos2πt0.01 s-x30 cm    …(1)

(c) Using

cos-θ=cosθin equation (1), we get:

y=1cos2πx30-t0.01

Velocity, v=dydt⇒v=2π0.01sin2πx30-t0.01And,Acceleration, a=dνdt⇒a=4π20.012cos2πx30-t0.01When x=50 cm, t=10 ms=10×10-3 s.Now,

v=2π0.01sin2π53-0.010.01  =2π0.01sin2π×23  =-2π0.01sin4π3  =- 200πsinπ3  =-200π×32  =-544 cm/s  =-5.4 m/sIn magnitude, v = 5.4 m/s.
Similarly,

a=4π20.012cos2π53-1  =4π2×104×12  ≈2×105 cm/s2 or 2 km/s2

Question 20:

A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of 160 N m−1 and is stretched by 1⋅0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring?

Answer:

Given,
Length of the string, L = 40 cm
Mass of the string = 10 gm
Mass per unit length

=1040=14 gm/cmSpring constant, k = 160 N/m

Deflection, x=1 cm                  =0.01 mTension, T= kx=160 ×0.01⇒T=1.6 N=16×104 dynNow, v=Tm=16×10414⇒v=8×102 cm/s=800 cm/s∴ Time taken by the pulse to reach the spring,

t=40800=120=0.05 s

Question 21:

Two blocks each having a mass of 3⋅2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB (figure 15-E5). The linear mass density of the wire AB is 10 g m−1 and that of CD is 8 g m−1. Find the speed of a transverse wave pulse produced in AB and CD.

Figure

Answer:

Given,
Mass of each block =

m1=m2=3.2 kgLinear mass density of wire AB = 10 gm−1 = 0.01 kgm−1
Linear mass density of wire CD = 8 gm−1 = 0.008 kgm−1
For string CD, velocity is defined as

v=Tm.
Here, T is the tension and m is the mass per unit length.
For string CD,

T=3.2×gThus, we have:

v =3.2×100.008    =32×1038    =2×1010    =20×3.14≈63 msFor string AB,

T=2×3.2g=64 NThus, we have:v=Tm  =640.01=6400  =80 m/s

Question 22:

In the arrangement shown in figure (15-E6), the string has a mass of 4⋅5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10 m s−2.

Figure

Answer:

Given,
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string:

m=4.5×10-32.25   =2×10-3 kg/mT=2g=20 NWave speed, ν=Tm                      =202×10-3                      = 104                       = 102 m/s=100 m/s
Time taken by the disturbance to reach the pulley:

t=sν =2100=0.02 s

Question 23:

A 4⋅0 kg block is suspended from the ceiling of an elevator through a string having a linear mass density of

19·2×10-3 kg m-1. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elevator accelerates up at the rate of 2⋅0 m s−2. Take g = 10 m s−2.

Answer:

Given,
Mass of the block = 4.0 kg
Linear mass density,

m=19.2×10-3 kg/mFrom the free body diagram,

T-4g-4a=0⇒T=4a+g      =42+10=48 N

Wave speed, ν=Tm                      =4819.2×103                      =2.5×10-3=50 m/s

Question 24:

A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of 60 cm s −1 on the string when the car is at rest and 62 cm s−1 when the car accelerates on a horizontal road. Find the acceleration of the car. Take g = 10 m s−2.

Answer:

 
Given,
Speed of the transverse pulse when the car is at rest, v1 = 60 cm s−1
Speed of the transverse pulse when the car accelerates, v2= 62 cm s−1
Let:
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Now,

Wave speed, ν=Tm=MgmWhen car is at rest:Tension in the string, T=Mg⇒v1=Mgm⇒Mgm=602    …(i)When car is having acceleration:
Tension,

T=Ma2+Mg2

Again, ν2=Tm⇒62=Ma2+Mg21/4m1/2

⇒ Ma2+Mg2m=622       …(ii)From equations (i) and (ii), we get:

Mgm×mMa2+Mg2=60622⇒ga2+g2=0.936⇒g2a2+g2=0.876⇒a2+100 0.876=100⇒ a2=12.40.867=14.15⇒a=3.76 m/s2Therefore, acceleration of the car is 3.76 m/s2.

Question 25:

A circular loop of string rotates about its axis on a frictionless horizontal place at a uniform rate so that the tangential speed of any particle of the string is

ν. If a small transverse disturbance is produced at a point of the loop, with what speed (relative to the string) will this disturbance travel on the string?

Answer:

Let,
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity

Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part at angle θ.
So,
Length of the element

=Rdθ, mass =mRdθCentrifugal force experienced by the element

=mRdθ ω2RResolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as

dF=2mR2dθω2 sinθ

Total force, F=∫0π/22mR2ω2sinθdθ                  =2mR2ω2 -cos θ                  =2mR2ω2And, 2T=2mR2ω2⇒T=mR2ω2Velocity of the transverse vibration is given as

V’=TmV’=mR2ω2m=ωRLinear velocity of the string, V =

ωR∴ Speed of the disturbance, V’ =  V

Question 26:

A heavy but uniform rope of length L is suspended from a ceiling. (a) Write the velocity of a transverse wave travelling on the string as a function of the distance from the lower end. (b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the ceiling? (c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse?

Answer:

(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as

v=Tm=mgxm⇒v=gx(b) Let the time taken be dt for the small displacement dx.
Thus, we have:

dt=dxv=dxgx

∴Total time, T=∫0Ldxgx=4Lg(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.

Now,

t=∫0ydxgx =4yg∴ Distance travelled by the particle in this time, S =

L-yUsing the equation of motion, we get:

S=ut+12 gt2⇒L-y=12 g×4yg2⇒L-y=2y⇒3y=L⇒y=L3Thus, the particle will meet the pulse at a distance

L3from the lower end.

Question 27:

Two long strings A and B, each having linear mass density

1·2×10-2 kg m-1, are stretched by different tensions 4⋅8 N and 7⋅5 N respectively and are kept parallel to each other with their left ends at x = 0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A?

Answer:

Given,
Linear density of each of two long strings A and B, m =

1.2×10-2 kg/mString A is stretched by tension Ta= 4.8 N.
String B is stretched by tension Tb= 7.5 N.
Let va and vb be the speeds of the waves in strings A and B.
Now,

va=Tam⇒va=4.81.2×10-2=20 m/svb=Tbm⇒vb=7.51.2×10-2=25 m/st1=0 in string At2=0+20 ms=20×10-3=0.02 sDistance travelled by the wave in 0.02 s in string A:
s

=20×0.02=0.4 mRelative speed between the wave in string A and the wave in string B, v’

=25-20=5 m/sTime taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m

t’=sv’=0.45=0.08 s

Question 28:

A transverse wave of amplitude 0⋅50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N. If the wave speed is 100 m s−1, what average power is the source transmitting to the wire?

Answer:

Given,
Amplitude of the transverse wave, r = 0.5 mm

=0.5×10-3 mFrequency, f = 100 Hz
Tension, T = 100 N
Wave speed, v = 100 m/s
Thus, we have:

ν=Tm⇒ν2=Tm⇒m=Tν2=1001002       =0.01 kg/mAverage power of the source:Pavg=2π2mνr2f2       =2 3.142 0.01×100×0.5×10-32×100       =2×9.86×0.25×10-6×104       =19.7×0.0025=0.049 W       =49×10-3 W=49 mW

Question 29:

A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density 6 g m−1 kept under a tension of 60 N. (a) Find the average power transmitted across a given point on the string. (b) Find the total energy associated with the wave in a 2⋅0 m long portion of the string.

Answer:

Given,
Frequency of the wave, f = 200 Hz
Amplitude, A = 1 mm = 10−3 m
Linear mass density, m = 6 gm−3
Applied tension, T = 60 N
Now,
Let the velocity of the wave be v.
Thus, we have:

v=Tm=606×10-3=102=100 m/s(a) Average power is given as

Paverage=2π2mνA2f2          =2×3.142×6×10-3×100×10-3×2002          =473×10-3=0.47 W(b) Length of the string = 2 m
Time required to cover this distance:

t=2100=0.02 sEnergy=Power×t            =0.47×0.02            =9.4×10-3 J=9.4 mJ

Question 30:

A tuning fork of frequency 440 Hz is attached to a long string of linear mass density 0⋅01 kg m−1 kept under a tension of 49 N. The fork produces transverse waves of amplitude 0⋅50 mm on the string. (a) Find the wave speed and the wavelength of the waves. (b) Find the maximum speed and acceleration of a particle of the string. (c) At what average rate is the tuning fork transmitting energy to the string?

Answer:

Given,
Frequency of the tuning fork, f = 440 Hz
Linear mass density, m = 0.01 kgm−1
Applied tension, T = 49 N
Amplitude of the transverse wave produce by the fork = 0.50 mm
Let the wavelength of the wave be

λ.

(a) The speed of the transverse wave is given by
ν=Tm

⇒v=490.01=70 m/s

Also, ν=fλ∴ λ=fv=70440=16 cm(b) Maximum speed (vmax) and maximum acceleration (amax):

We have: y=A sin ωt-kx

∴ ν=dydt=Aω cos ωt-kxNow,νmax=dydt=Aω        =0.50×10-3×2π×440        =1.3816 m/s.And, a=d2ydt2⇒a=-Aω2 sin ωt-kxamax=-Aω2       =0.50×10-3×4π2 4402       =3.8 km/s2(c) Average rate (p) is given by

p=2π2νA2f2  =2×10×0.01×70×0.5×10-32×4402  =0.67 W

Question 31:

Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90°, what is the resultant amplitude?

Answer:

Given,
Phase difference between the two waves travelling in the same direction,

ϕ=90°=π2.
Frequency f and wavelength

λare the same. Therefore, ω will be the same.
Let the wave equations of two waves be:

y1=rsinωt

y2=rsinωt+π2Here, r is the amplitude.
From the principle of superposition, we get:

y=y1+y2  =rsinωt+rsinωt+π2  =r sinωt+sinωt+π2  =r2sinωt+ωt+π22cosωt-ωt-π22  =2rsinωt+π4cos-π4  =2rsinωt+π4∴ Resultant amplitude,

r’=2r=42 mm

Question 32:

Figure (15-E7) shows two wave pulses at t = 0 travelling on a string in opposite directions with the same wave speed 50 cm s−1. Sketch the shape of the string at t = 4 ms, 6 ms, 8 ms, and 12 ms.

Figure

Answer:


Given,
Speed of the wave pulse travelling in the opposite direction, v = 50 cm s−1 = 500 mm s−1
Distances travelled by the pulses:
Using s = vt, we get:

In t=4 ms=4×10-3 s,s=νt=500×4×10-3=2 mm.In t=6 ms=6×10-3 s,s=500×6×10-3=3 mm.In t=8 ms=8×10-3 s,s=νt=500×8×10-3= 4 mm.In t=12 ms=12×10-3 s,s=500×12×10-3=6 mm.The shapes of the string at different times are shown in the figure.

Page No 326:

Question 33:

Two waves, each having a frequency of 100 Hz and a wavelength of 2⋅0 cm, are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced 0⋅015 s later than the first one at the same place, (b) if the two waves were produced at the same instant but first one was produced a distance 4⋅0 cm behind the second one? (c) If each of the waves has an amplitude of 2⋅0 mm, what would be the amplitudes of the resultant waves in part (a) and (b) ?

Answer:

Given:
Two waves have same frequency (f), which is 100 Hz.
Wavelength (λ) = 2.0 cm

=2×10-2 m

Wave speed, ν =f×λ=100×2×10-2 m/s                           =2 m/s(a) First wave will travel the distance in 0.015 s.

⇒x=0.015×2  =0.03 mThis will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:

ϕ=2πxλ   =2π2×10-2×0.03=3π(b) Path difference between the two waves, x = 4 cm = 0.04 m
So, the corresponding phase difference will be as follows:

⇒ϕ=2πxλ         =2π2×10-2×0.04        =4π(c) The waves have same frequency, same wavelength and same amplitude.
Let the wave equation for the two waves be as follows:

y1=r sin ωtAnd, y2=r sin ωt+ϕBy using the principle of superposition:y=y1+y2  =rsinωt+ωt+ϕ   =2rsinωt+ϕ2 cosϕ2∴ Resultant amplitude

=2r cosϕ2

So, when ϕ=3x:⇒r=2×10-3 mRresultant=2×2×10-3 cos 3π2            =0

Again, when ϕ=4π:Rresultant=2×2×10-3 cos 4π2            =4×10-3×1             =4 mm

Question 34:

If the speed of a transverse wave on a stretched string of length 1 m is 60 m−1, what is the fundamental frequency of vibration?

Answer:

Length of a stretched string (L) = 1 m
Wave speed (v) = 60 m/s
Fundamental frequency (f0) of vibration is given as follows:

f0=v2L

=602×1=30 s-1=30 Hz

Question 35:

A wire of length 2⋅00 m is stretched to a tension of 160 N. If the fundamental frequency of vibration is 100 Hz, find its linear mass density.

Answer:

Given:
Length of the wire (L)= 2.00 m
Fundamental frequency of the vibration (f0) = 100 Hz
Applied tension (T) = 160 N

Fundamental frequency, f0=12LTm⇒10=14160m⇒m=1×10-3 kg/m⇒m=1 g/mSo, the linear mass density of the wire is 1 g/m.

Question 36:

A steel wire of mass 4⋅0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. Find the frequency and wavelength of the fourth harmonic of the fundamental.

Answer:

Given:
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m)

=480 g/cm =0.005 kg/m Wave speed, ν=Tm  =500.005=100 m/s

Fundamental frequency, fo=12LTm   =12×0.8×500.005   =1002×0.8=62.5 HzFirst harmonic=62.5 HzIf f4= frequency of the fourth harmonic:⇒f4=4f0=62.5×4⇒f4=250 HzWavelength of the fourth harmonic, λ4=νf4=100250⇒λ4=0.4 m=40 cm

Question 37:

A piano wire weighing 6⋅00 g and having a length of 90⋅0 cm emits a fundamental frequency corresponding to the “Middle C”

ν=261·63 Hz. Find the tension in the wire.

Answer:

Given:
Length of the piano wire (L)= 90.0 cm = 0.90 m
Mass of the wire = 6.00 g = 0.006 kg
Fundamental frequency (fo) = 261.63 Hz

Linear mass density, m=690 gm/cm     =6×10-390×10-2 kg/m     =6900 kg/m Fundamental frequency, fo=12LTm

⇒ 261.63=12×0.09 T×9006⇒0.18×261.63=150 T⇒150 T=261.63×0.182⇒T=261.63×0.182150      =1478.52 N≈1480 NHence, the tension in the piano wire is 1480 N.

Question 38:

A sonometer wire having a length of 1⋅50 m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire?

Answer:

Given:
Length of the sonometer wire (L) = 1.50 m
Let the

first harmonic be f0 and the second harmonic be f1.According to the question,

f1 =256 Hz

1st harmonic for fundamental frequency, f0=f12=2562=128 Hz
When the fundamental wave is produced, we have:

λ2=1.5 m⇒ λ=3 m

Wave speed, v=f0λ⇒v=128×3=384 m/s

Question 39:

The length of the wire shown in figure (15-E8) between the pulley is 1⋅5 m and its mass is 12⋅0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

Figure

Answer:

Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm

Mass per unit length, m=121.5 g/m  =8×10-3 kg/m Tension in the wire, T=9×g                                  =90 N
Fundamental frequency is given by:

f0=12L TmFor second harmonic (when two loops are produced):

f1=2f0=11.5 908×10-3           =106.061.5           =70.7 Hz≈70 Hz

Question 40:

A one-metre long stretched string having a mass of 40 g is attached to a tuning fork. The fork vibrates at 128 Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops?

Answer:

Given:
Length of the stretched string (L) = 1.00 m
Mass of the string =40 g
String is attached to the tuning fork that vibrates at the frequency (f) = 128 Hz
Linear mass density (m)

=40×10-3 kg/mNo. of loops formed, (n) = 4

L=nλ2⇒λ=2Ln=2×14⇒λ=0.5 mWave speed (v)=fλ=128×0.5⇒v=64 m/s We know: v=Tm⇒T=ν2m      =642×40×10-3      =163.84≈164 NHence, the tension in the string if it is to vibrate in four loops is 164 N.

Question 41:

A wire, fixed at both ends is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz. (a) What could be the maximum value of the fundamental frequency? (b) If transverse waves can travel on this string at a speed of 40 m s−1, what is its length?

Answer:

Given:
Wire makes a resonant frequency of 240 Hz and 320 Hz when its both ends are fixed.
Therefore, fundamental frequency (f0) of the wire must be the factor of 240 Hz and 320 Hz.
(a) Maximum value of fundamental frequency, f0 = 80 Hz

(b) Wave speed (v) = 40 m/s
And if

λ is the wavelength:

λ2=L∴ v=λ×f0⇒v=2×L×f0⇒L=402×80⇒L=14 m=0.25 m

Question 42:

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2⋅0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1⋅6 cm. Find the length of the string.

Answer:

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be ‘n‘ loops and

λbe the wavelength.

λ=

2×Separation between the consecutive nodes

λ1=2×2=4 cm

λ2=2×1.6=3.2 cmLength of the wire is L.

In the first case:

L=nλ12In the second case:

L=n+1λ22⇒nλ12=n+1 λ22⇒n×4=n+13.2⇒4n-3.2n=3.2⇒0.8 n=3.2⇒n=4∴ Length of the string,

L=nλ12=4×42=8 cm

Question 43:

A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is 220 m s−1 and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0⋅5 cm, write a suitable equation describing the motion.

Answer:

Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s

Wavelength, λ=vf=220660=13 m(a) No. of loops, n = 3

L=n2λ

⇒L=32×13⇒L=12 m=50 cm(b) Equation of resultant stationary wave can be given by:

y=2Acos2πxλsin2πvLλ⇒y=0.5 cos2πx13sin2π×220×t13⇒y=0.5 cm cos6πx m-1 sin1320πt s-1

Question 44:

A particular guitar wire is 30⋅0 cm long and vibrates at a frequency of 196 Hz when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes?

Answer:

Given:
Length of the guitar wire (L1) = 30.0 cm = 0.30 m
Frequency, when no finger is placed on it, (f1) =196 Hz
And (f2) =220 Hz, (f3) = 247 Hz, (f4) = 262 Hz and (f5) = 294 Hz
The velocity is constant for a medium.
We have:

f∝1L

⇒f1f2=L2L1⇒196220=L20.3⇒L2=196×0.3220=0.267 m⇒L2=26.7 cmAgain,

f3=247 Hz

⇒f3f1=L1L3⇒247196=0.3L3⇒L3=196×0.3247=0.238 m⇒L3=23.8 cmSimilarly, L4=196×0.3262=0.224 m⇒L4=22.4 cmAnd, L5=20 cm

Question 45:

A steel wire fixed at both ends has a fundamental frequency of 200 Hz. A person can hear sound of maximum frequency 14 kHz. What is the highest harmonic that can be played on this string which is audible to the person?

Answer:

Given:
Fundamental frequency (f0) of the steel wire = 200 Hz
Let the highest harmonic audible to the person be n.

Frequency of the highest harmonic, f‘ = 14000 Hz
f’=nf0    …(1)

f’f0=14000200⇒nf0f0=70⇒ n=70Thus, the highest harmonic audible to man is the 70th harmonic.

Question 46:

Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string. (b) Which harmonics of the fundamental are the given frequencies? (c) Which overtones are these frequencies? (d) If the length of the string is 80 cm, what would be the speed of a transverse wave on this string?

Answer:

Given:
Let the three resonant frequencies of a string be

f1=90 Hzf2=150 Hzf3=210 Hz(a) So, the highest possible fundamental frequency of the string is

f=30 Hz, because f1, f2 and f3 are the integral multiples of 30 Hz.

(b) So, these frequencies can be written as follows:

f1=3ff2=5ff3=7fHence, f1, f2, and f3 are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are f, 2f, 3f, 4f, 5f
∴ 3f = 2nd overtone and 3rd harmonic
5f = 4th overtone and 5th harmonic
7th= 6th overtone and 7th harmonic

(d) Length of the string (L) = 80 cm = 0.8 m
Let the speed of the wave be v.
So, the frequency of the third harmonic is given by:

f1=32×L v⇒90=32×80×v⇒v=90×2×803      =30×2×80      =4800 cm/s⇒v=48 m/s

Question 47:

Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2 : 1 the radii are in the ratio 3 : 1 and the densities are in the ratio 1 : 2. Find the ratio of their fundamental frequencies.

Answer:

Given:
The tensions in the two wires are in the ratio of 2:1.

⇒T1T2=2Ratio of the radii is 3:1.

⇒r1r2=3=D1D2Density in the ratios of 1:2.

⇒ρ1ρ2=12Let the length of the wire be L.

Frequency, f=1LDTπρ

⇒f1=1LD1T1πρ1⇒f2=1LD2T2πρ2

∴ f1f2=LD2LD1T1T2πρ2πρ1⇒f1f2=1321×21⇒ f1:f2=2:3

Question 48:

A uniform horizontal rod of length 40 cm and mass 1⋅2 kg is supported by two identical wires as shown in figure (15-E9). Where should a mass of 4⋅8 kg be placed on the rod so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone? Take g = 10 m s−2.

Figure

Answer:

Given:
Length of the rod (L) = 40 cm = 0.40 m
Mass of the rod (m) = 1.2 kg
Let the mass of 4.8 kg be placed at x distance from the left.
As per the question, frequency on the left side = f0
Frequency on the right side = 2f0
Let tension be T1 and T2 on the left and the right side, respectively.
∴ 12LT1m=22LT2m⇒T1T2=2⇒T1T2=4    … (1)From the free body diagram:

T1+T2=48+12=60 N⇒ 4T2+T2=5T2=60 N   using equation 1∴T2=12 Nand T1=48 NNow, taking moment about point A:

T2×0.4=48x+12×0.2⇒4.8=48x-2.4⇒4.8x=2.4⇒x=2.44.8=120 m=5 cmTherefore, the mass should be placed at a distance of 5 cm from the left end.

Question 49:

Figure (15-E10) shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is 1⋅0 mm2 and that of the aluminium wire is 3⋅0 mm2. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminium is 2⋅6 g cm−3 and that of steel is 7⋅8 g cm−3.

Figure

Answer:

Given:
Length of the aluminium wire (La)= 60 cm = 0.60 m
Length of the steel wire (Ls)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (Aa)  = 1.0 mm2
Area of cross-section of the steel wire (As) = 3.0 mm2
Density of aluminium (ρa) = 2⋅6 g cm−3
Density of steel  (ρs) = 7⋅8 g cm−3

Mass per unit length of the steel, ms =ρs×As                                                              =7.8×10-2 gm/cm                                                              =7.8×10-3 kg/mMass per unit length of the aluminium, mA=ρAAA                                                                       =2.6×10-2×3 gm/cm                                                                       =7.8×10-2 gm/cm                                                                       =7.8×10-3 kg/m  
A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.

⇒v=Tm     =407.8×10-3-      =4×1047.8⇒v=71.6 m/sFor minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm

⇒ Wavelength, λ=2×20=40 cmOr λ=0.4 m∴Frequency, f=vλ=71.60.4=180 Hz

Question 50:

A string of length L fixed at both ends vibrates in its fundamental mode at a frequency

νand a maximum amplitude A. (a) Find the wavelength and the wave number k. (b). Take the origin at one end of the string and the X-axis along the string. Take the Y-axis along the direction of the displacement. Take t = 0 at the instant when the middle point of the string passes through its mean position and is going towards the positive y-direction. Write the equation describing the standing wave.

Answer:

Given:
Length of the string = L
Velocity of wave is given as:

V=TmIn fundamental mode, frequency = ν

⇒ν=12LTm(a)

Wavelength, λ=VelocityFrequency

⇒λ=Tm12LTm=2LWave number, K=2πλ=2π2L=πL(b) Equation of the stationary wave:

y=Acos2πxλsin2πVtλ      =Acos2πx2Lsin2πVt2L   ∵ ν=V2L      =AcosπxLsin2πνt

Page No 327:

Question 51:

A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s−1 and the amplitude is 0⋅5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end and  t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.

Answer:

Given:
Length of the string (L) = 2.0 m
Wave speed on the string in its first overtone (v) = 200 m/s
Amplitude (A) = 0.5 cm

(a) Wavelength and frequency of the string when it is vibrating in its 1st overtone (n = 2):

L=nλ2

⇒λ=L=2 m⇒f=νλ=2002=100 Hz(b) The stationary wave equation is given by:

y=2A cos2πxλsin2πvtλ  =0.5cos2πx2sin2π×200 t2  =0.5 cmcosπm-1 xsin200πs-1 t

Question 52:

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by

y=0·4 cm sin0·314 cm-1 x cos 600π s-1 t.

(a) What is the frequency of vibration? (b) What are the positions of the nodes? (c) What is the length of the string? (d) What is the wavelength and the speed of two travelling waves that can interfere to give this vibration?

Answer:

Given:
The stationary wave equation of a string vibrating in its third harmonic is given by
y = (0.4 cm) sin [(0.314 cm−1) x]cos [(.600 πs−1) t]
By comparing with standard equation,

y = A sin (kx) cos (wt)(a) From the above equation, we can infer the following:
ω=600 π

⇒2πf=600 π⇒f=300 HzWavelength,

λ=2π0.314=2×3.140.314

⇒λ=20 cm(b) Therefore, the nodes are located at 0cm, 10 cm, 20 cm, 30 cm.

(c) Length of the string, l =

nλ2

⇒l=3λ2=3×202=30 cm(d)

y=0.4 sin 0.314 x cos 600 πt

=0.4sinπ10 xcos600πt λand

νare the wavelength and velocity of the waves that interfere to give this vibration.

λ=20 cmν=ωk=600 ππ10=6000 cm/s⇒ν=60 m/s

Question 53:

The equation of a standing wave, produced on a string fixed at both ends, is

y=0·4 cm sin 0·314 cm-1 x cos 600π s-1 t.

What could be the smallest length of the string?

Answer:

Given:
Equation of the standing wave:

y=0.4 cm sin 0.314 cm-1 xcos 600 πs-1 t⇒k=0.314=π10Also, k=2πλ⇒λ=20 cmWe know:

L=nλ2For the smallest length, putting n = 1:

⇒L=λ2=20 cm2=10 cmTherefore, the required length of the string is 10 cm.

Question 54:

A 40 cm wire having a mass of 3⋅2 g is stretched between two fixed supports 40⋅05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is 1⋅0 mm2, find its Young modulus.

Answer:

Given:
Length of the wire (L) = 40 cm = 0.40 m
Mass of the wire = 3.2 g = 0.003 kg
Distance between the two fixed supports of the wire = 40.05 cm
Fundamental mode frequency = 220 Hz
Therefore, linear mass density of the wire (m) is given by:

m=0.00320.4=8×10-3 kg/mChange in length, ∆L=40.05-40=0.05 cm       =0.05×10-2 mStrain=∆LL=0.05×10-20.4           =0.125×10-2f0=12LTm    =12×0.4005 T8×10-3 ⇒220×220=10.8012×T×1038⇒T×1000=220×220×0.641×0.8⇒T=248.19 NStress=TensionArea=248.191 mm2=248.1910-6⇒Stress=248.19×106Young’s modulus, Y=stressstrain                                  =248.19×1060.125×10-2⇒Y=19852×108        =1.985×1011 N/m2Hence, the required Young’s modulus of the wire is

1.985×1011 N/m2.

Question 55:

Figure (15-E11) shows a string stretched by a block going over a pulley. The string vibrates in its tenth harmonic in unison with a particular tuning for. When a beaker containing water is brought under the block so that the block is completely dipped into the beaker, the string vibrates in its eleventh harmonic. Find the density of the material of the block.

Figure

Answer:

Density of the block = ρ
Volume of block = V
∴ Weight of the block is, W = ρVg
∴ Tension in the string, T = W

The tuning fork resonates with different frequencies in the two cases.
Let the tenth harmonic be f10.

f10=102LTm    =102L ρ VgmHere, m is the mass per unit length of the string and L is the length of the string.

When the block is immersed in water (density = ρ ), let the eleventh harmonic be f11.

f11=112LT’m     =112Lρ-ρw VgmThe frequency (f) of the tuning fork is same.

∴ f10=f11⇒102LρVgm=112Lρ-ρω Vgm⇒100121=ρ-1ρ   because ρω=1  gm/cc⇒100 ρ=121 ρ-121⇒ρ=12121=5.8 gm/cc        =5.8×103 kg/m3Therefore, the required density is

5.8×103 kg/m3.

Question 56:

A 2⋅00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

Answer:

Given:
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density, m

=0.082=0.04 kg/m

Tension, T=256 NWave velocity, v=Tm⇒v=2560.04=1602⇒v=80 m/sFor fundamental frequency:

L=λ4⇒λ=4L=4×2=8 m⇒f=vλ=808=10 Hz(a) The frequency overtones are given below:

1st overtone=3f=30 Hz2nd overtone=5f=50 Hz(b)

λ=4l=4×2=8 m

∴ λ1=vf1=8030=2.67 m λ2=vf2=8050=1.6 mHence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.

Question 57:

A heavy string is tied at one end to a movable support and to a light thread at the other end as shown in figure (15-E12). The thread goes over a fixed pulley and supports a weight to produce a tension. The lowest frequency with which the heavy string resonates is 120 Hz. If the movable support is pushed to the right by 10 cm so that the joint is placed on the pulley, what will be the minimum frequency at which the heavy string can resonate?

Figure

Answer:

Let T be the tension in the string and m be the mass per unit length of the heavy string.
In the first part of the question, the heavy string is fixed at only one end.
So, the lowest frequency is given by:

f0=14LTm    …1When the movable support is pushed by 10 cm to the right, the joint is placed on the pulley and the heavy string becomes fixed at both the ends (keeping T and m same).
Now, the lowest frequency is given by:

f0’=12LTm    …2Dividing equation (2) by equation (1), we get:

f0’=2f0=240 Hz

HC Verma Solutions for Class 11 Physics – Part 1

HC Verma Solutions for Class 12 Physics – Part 2

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