1st PUC Physics Question Bank Chapter 15 Waves

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Karnataka 1st PUC Physics Question Bank Chapter 15 Waves

Question 1.
What is a wave?
Answer:
The disturbance produced in a medium is called wave.

Question 2.
What is wave motion?
Answer:
The propagation of disturbance from one point to another is called wave motion.

Question 3.
What happens to the particles of the medium when a wave travels through the medium?
Answer:
Particles of the medium execute periodic motion about its mean position.

Question 4.
What is that moves in a wave?
Answer:
Energy is transported from one particle to another in the form of wave.

Question 5.
What are mechanical waves? Give examples.
Answer:
The waves that require a material medium for propagation are called mechanical waves.
Example: sound waves, water waves, waves in a string.

Question 6.
Name the essential condition for the propagation of mechanical waves.
Answer:
The elastic and inertial properties of the medium are required for the propagation of waves through the medium.

Question 7.
How mechanical waves are produced?
Answer:
Mechanical waves are produced due to the periodic motion executed by the particles of the medium.

Question 8.
What are electromagnetic waves?
Answer:
The waves that does not require a material medium for propagation are called electromagnetic waves. They can travel through vacuum also.

Question 9.
Give few examples for electromagnetic waves.
Answer:
Light waves, X-rays, radio waves etc.

Question 10.
Can electromagnetic wave travel through a material medium?
Answer:
Electromagnetic wave can travel through a material medium also.

Question 11.
What are the speed of the electromagnetic waves through vacuum.
Answer:
In vacuum all electromagnetic waves have same speed i.e., c = 2.99792458 × 10⁸ms^-1 ≅ 3 × 10⁸ ms^-1

Question 12.
What are matter waves?
Answer:
The waves associated with material particles in motion are called matter waves.

Question 13.
Name the device which is based on matter waves.
Answer:
Electron microscope.

Question 14.
Mention the names of famous scientist associated with wave motion.
Answer:
Some of the famous scientists associated with physics of wave motion are Christian Huygens, Robert Hooke, Issac Newton etc.

Question 15.
What are longitudinal waves?
Answer:
The wave is longitudinal, if the particles of the medium oscillate along the direction of propagation of the wave.

Question 16.
Give an example for longitudinal wave.
Answer:
Sound wave

Question 17.
Explain the formation of longitudinal waves with an example.
Answer:
Consider a long pipe filled with air and piston at one end. If the piston is pushed forward and pulled back, alternate condensation (compression) (region of higher density) and rarefaction (region of lower density) is produced in the medium air. If the push-pull of the piston is continuous and periodic, a sinusoidal wave travels along the length of the pipe. Since the air molecules vibrate along the direction of propagation of wave, the wave is longitudinal.

Question 18.
What are condensation (compression) and rarefaction in a longitudinal wave?
Answer:
In a medium the region of higher density is called condensation (compression) and a region of lower density is called rarefaction.

Question 19.
What are transverse waves?
Answer:
The wave is transverse, if the particles of the medium oscillate perpendicular to the direction of propagation of the wave.

Question 20.
Explain the formation of transverse wave in a stretched string.
Answer:
Consider a string with its one end fixed to a rigid support. If the free end is moved up and down once, a wave pulse travel along the string towards the fixed .end. If the free end of the string is moved up and down continuously and periodically, a sinusoidal wave travel along the string. During this, the elements of the string vibrate perpendicular to the direction of propagation of wave. Hence the wave is transverse in nature.

Question 21.
Distinguish between longitudinal and transverse waves.
Answer:

Longitudinal wave	Transverse wave
1. The particles of the medium oscillate along the direction of propagation of the wave.	1. The particles of the medium oscillate perpendicular to the direction of propagation of the wave.
2. Longitudinal wave travels in the form of alternate compressions and rare factions.	2. Transverse wave travels in the form of alternate crests and troughs.
3. Travels through solids and fluids.	3. Travels through solids and but not in fluids.

(ii) The sinusoidal wave travelling along negative x-axis is
y (x, t) = a sin(kx + ωt + Φ)
where y(x,t) is the displacement as a function of position x and time t.
a is the amplitude of the wave
ω is the angular frequency
Φ is the initial phase.

Question 22.
Define amplitude of a wave.
Answer:
It is the maximum displacement of a particle of the medium from its equilibrium position.

Question 23.
Define phase of a wave.
Answer:
Phase of a wave at any instant of time is the state of vibration of the particle of the medium at that instant of time.

Question 24.
Define wavelength of a wave.
Answer:
It is the minimum distance between two points having the same phase.
or
It is the minimum distance between two consecutive crests or troughs in a wave.

Question 25.
Define propagation constant or wave number.
Answer:
It is defined as the number of waves present in unit length of the medium.

Question 26.
Define period of a wave.
Answer:
It is the time during which the particles of the medium completes one complete oscillation, or It is the time during which one complete wave is set up in the medium.

Question 27.
Define frequency of a wave.
Answer:
It is the number oscillations executed by the particles of the medium in one second.

Question 28.
Mention the relation between time period and frequency.
Answer:
Period = \(\frac{1}{\text { frequency }}\) or frequency = \(\frac{1}{\text { Period }}\)
i.e., T = \(\frac{1}{v}\) or v = \(\frac{1}{T}\)

Question 29.
Obtain the relation between wavelength (λ) and wave number (k).
Answer:
The displacement of a wave is given by y(x, t) = A sin (kx – ωt + Φ)
When Φ = 0 and t = 0,
y(x,0) = A sin kx
since sine function repeats its value after every 2π radian.
sin kx = sin(kx + 2nπ)
= sink (\(x+\frac{2 n \pi}{k}\))
Thus the displacement at x and x + [/latex]\frac{2 n \pi}{k}[/latex] is same.
∵the least distance between two points which are in phase is wavelength.
∴λ = \(\frac{2 \pi}{k}\) [∵ n = 1 for least distance]
k = \(\frac{2 \pi}{\lambda}\)

Question 30.
Mention the unit of angular wave number.
Answer:
The unit of angular wave number is radian per metre (rad m^-1).

Question 31.
Give the relation between angular frequency and frequency of the wave and mention its unit in SI system.
Answer:
Angular frequency w = 2πv
or ω = \(\frac{2 \pi}{T}\)
The unit of angular frequency is radian per second (rad s^-1)

Question 32.
Obtain the relation between angular frequency and time period.
Answer:
The displacement of a wave at any instant t is .
y(x,t) = asin(kx – ωt + Φ)
when x = 0 and Φ = 0,
y (0, t) = a sin(-ωt) = -a sinωt
sine function repeats its value after one complete revolution i.e., after T seconds
-a sinωt = -a sinω(t + T)
= -a sin(ωt + ωT)
∵since sine function repeats its value after every 2π radian.
ωT = 2π
ω = \(\frac{2 \pi}{T}\)
further \(\frac{1}{T}\) = v
∴ω = 2πv

Question 33.
Define speed of a wave.
Answer:
It is the distance travelled by the wave in one second.

Question 34.
Obtain an expression for the speed of a travelling wave.
Answer:
Consider a wave travelling along + x-direction. Let P be a point on the crest of the wave. As the wave travels, the reference point also moves. Let the point moves through a distance Δx in a time Δt then speed of the wave
v = \(\left(\frac{\Delta x}{\Delta t}\right)\)

Since the point P always remains at crest phase at ‘t’ = phase at (t + ∆t)
kx – ωt = k(x + ∆x) – ω(t + ∆t)
kx – ωt = kx + k. ∆x – ωt -ω∆t
k∆x – ω∆t = 0
if ∆x and ∆t are very small,
\(\frac{d x}{d t}=\frac{\omega}{k}=v\)

Question 35.
Obtain a relation between velocity frequency and wavelength of a wave.
Answer:
We know, velocity of the wave v = \(\frac{\omega}{k}\)
Multiply and dividing by 2π

v = vλ

Question 36.
Obtain an expression for speed of a transverse wave in a stretched string.
Answer:
The speed of transverse wave in a stretched string is

• Directly proportional to the restoring force
• Inversely proportional to the inertial properties of the medium.

For a stretched string, the tension T provides the restoring force. The inertial properties is linear mass density μ.
The dimensions of tension T is [MLT^-2]
and the dimensions of linear mass density μ = \(\frac{\text { mass }}{\text { length }}\) = [ML^-1]
\(\frac{T}{\mu}=\frac{\left[M L T^{-2}\right]}{\left[M L^{-1}\right]}\) = [L²T^-2] = [LT^-1]² = v²
∴v ∝ \(\sqrt{\frac{T}{\mu}}\)
v = c\(\sqrt{\frac{T}{\mu}}\)
It can be shown that the constant C = 1
∴ v = \(\sqrt{\frac{T}{\mu}}\)

Question 37.
Transverse waves are possible only in solids but not in fluids. Why?
Answer:
Solids have shearing stress i.e., they sustain shearing stress. Fluids have no shape on their own and they yield to shearing stress. Hence transverse wave and possible only solid but not in fluids.

Question 38.
Obtain an expression for speed of longitudinal waves (sound waves) in a gaseous medium.
Answer:
When a longitudinal wave travels, particles of the medium oscillate along the direction of propagation of the wave forming alternate compressions and rarefactions.
The speed of the wave depends on elastic and inertial properties of the medium.
The elastic property for the gaseous medium is the bulk modulus B.
B=\(\frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)}\)
where ∆P is change in pressure and \(\left(\frac{\Delta V}{V}\right)\) is the volume of the strain.
the inertial property for the propagation of the wave is mass density (ρ)
velocity of the wave depends on \(\left(\frac{B}{\rho}\right)\)
dimensions of \(\left(\frac{B}{\rho}\right)\) =\(\frac{\left[M L^{-1} T^{-2}\right]}{\left[M L^{-3}\right]}\) = [L²T^-2] = [LT^-1]² = v²
∴v ∝\(\sqrt{\frac{B}{\rho}}\)
v = c\(\sqrt{\frac{B}{\rho}}\)
c is a dimensionless constant. It can be shown that C = 1
v = \(\sqrt{\frac{B}{\rho}}\)

Question 39.
Mention the expression for speed of sound in a solid medium.
Answer:
The speed sound, v = \(\sqrt{\frac{Y}{\rho}}\)
where Y is the Young’s modulus of elasticity.
ρ density of the medium. .

Question 40.
Explain Newton’s formula for the speed of sound in air.
Answer:
The speed of sound (longitudinal waves) in air (gaseous medium) is
v = \(\sqrt{\frac{Y}{\rho}}\) ………(1)
where B is the bulk modulus
ρ is the density of the medium.
B = \(\frac{\text { stress }}{\text { volume strain }}=-\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)}\) ….(2)
When sound travels through a gaseous medium, alternate compressions and rarefactions are produced.
In a compressed region, volume decreases and hence pressure increases. In a rarefied region, the volume increases and hence pressure decreases.
Newton assumed that these changes in pressure and volume takes place under isothermal condition i.e., at constant temperature.
∴ PV = constant ……(3)
Differentiating equation (3)
P(∆V) + (∆P)V = 0
P(∆V) = -(∆P)V
P= \(-\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)}\) ……(4)
and comparing (2) and (4)
B = P
Substituting B = P in equation (1)
∴speed of sound v = \(\sqrt{\frac{P}{\rho}}\)
This equation is known as Newton’s formula for the speed of sound.

Question 41.
Why Newton’s formula for the speed of sound in a gaseous medium is incorrect?
Answer:
The experimental value of speed of sound in air at NTP is 331 ms^-1.
But according to Newton’s formula, the speed of sound in air is
v = \(\sqrt{\frac{P}{\rho}}\)
at NTP, P = 1.013 × 10⁵ Nm^-2
ρ = 1.29 kg m^-3
v = \(\sqrt{\frac{1 \cdot 013 \times 10^{5}}{1 \cdot 29}}\) = 280 ms^-1
∴Since there is a large difference 51 ms^-1 or 15% error between experimental value and theoretical value, Newton’s formula requires a correction.

Question 42.
Explain Newton – Laplace formula for speed of sound in air (gaseous medium)
Answer:
According to Newton, the changes in pressure during the propagation of sound is isothermal. But according to Laplace, in a compressed region temperature increases and in a rarefied region the temperature decreases. Further the pressure changes take place rapidly and there is no time for heat to flow from compressed region to rarefied region to maintain constant temperature. Hence changes takes place at adiabatic conditions.
For adiabatic condition,
PV^γ = constant
taking long on both sides,
log P + γlog V = constant
differentiating,

from (1) and (2) B = γP
∴Speed of sound in a gaseous medium (air) is
v = \(\sqrt{\frac{B}{\rho}}=\sqrt{\frac{\gamma P}{\rho}}\)
This formula is known as Newton’s Laplace formula.
where y = \(\frac{C_{P}}{C_{V}}\) ratio of specific heats
P = pressure
ρ = density of the medium.

Question 43.
What is a wave pulse?
Answer:
A single disturbance produced in a medium is called wave pulse.

Question 44.
What is a continuous wave?
Answer:
A series of identical wave pulses generated one after the other successively is called a continuous wave.

Question 45.
What is a progressive wave?
Answer:
A wave which travels continuously in a medium in the same direction is called progressive wave.

Question 46.
Mention the equation for a progressive wave travelling along (i) positives x-direction (ii) negative x-direction and explain the terms used.
Answer:
(i) A sinusoidal wave travelling along positive x-axis is
y(x,t) = asin(kx – ωt + Φ)

(ii) Sinusoidal wave travelling along negative x – axis is
y(x, t) = a sin(kx + ωt + Φ)
where k is the propagation constant,
ω is the angular frequency.
Φ is the initial phase angle.

Question 47.
Transverse waves are not produced in fluids. Why?
Answer:
The transverse wave propagates in the form of crests and troughs which involves the change in the shape of the medium. Since fluids do not posses the property of elasticity, they do not have the shape of their own. Hence transverse wave cannot be produced in fluids.

Question 48.
Why are the longitudinal waves considered as pressure waves.
Answer:
Longitudinal waves travel in the form of compressions and rarefactions. Compression is a region of higher density where the pressure is maximum rarefaction is a region of lower density where the pressure is minimum. Hence longitudinal waves are called pressure waves.

Question 49.
State and explain the principle of superposition.
Answer:
When two or more wave pulses overlap, the resultant displacement is equal to the algebraic sum of the displacements produced by the individual wave pulses.
Let y₁ and y₂ are the displacements produced by wave pulses at a point in a medium. When they overlap the resultant displacement y is given by
y = y₁ + y₂

Question 50.
What is the basic principle behind the phenomenon of superposition.
Answer:
The principle of superposition is the basic to the phenomenon of interference.

Question 51.
Using the principle of superposition explain constructive and destructive interference.
Answer:
Consider two identical waves of angular frequency ω, wave number k and wavelength λ travelling along
positive x-direcdon. Let Φ be the phase difference between the two waves.
The two waves represented as
y₁ = asin (kx – ωt) ……(1)
y₂ = a sin (kx – ωt + Φ) …..(2)
using the principle of superposition, the resultant displacement y is
y = y₁ + y₂
y = a sin(kx – ωt) + a sin(kx – ωt + Φ)
y = a {sin(kx – ωt) + sin(kx – ωt + Φ)}

The above equation is also simple harmonic travelling wave in the positive x-direction with the same angular
frequency ω and wavelength λ but with an initial phase angle \(\left(\frac{\phi}{2}\right)\)
In equation (1), 2a cos \(\left(\frac{\phi}{2}\right)\) represents amplitudes of the resultant wave.
When Φ = 0, i.e., when two waves are in phase, the resultant amplitude is maximum i.e., 2A. Then two waves are said to interfere constructively.
When Φ = π, i.e., when two waves are out of phase, the resultant amplitude is minimum, i.e., 0. Then two waves are said to interfere destructively.

Question 52.
What happens to a wave, if it meets a rigid boundry?
Answer:
The wave gets reflected.

Question 53.
What happens to a wave, if it meets a boundry which is not completely rigid?
Answer:
A part of the incident wave is reflected and a part is transmitted into the second medium.

Question 54.
What is the phase difference between the incident and the reflected wave at a rigid boundry?
Answer:
The phase difference between the incident and the reflected wave is n radian or 180°.

Question 55.
What is the phase difference between incident wave and the reflected wave at a open boundry?
Answer:
There is no phase difference incident and the reflected wave i.e., the phase difference is zero.

Question 56.
A wave pulse y =asin(kx – ωt) is incident on (i) rigid boundry (ii) open boundry. What is the reflected wave in each case.
Answer:
(i) When a wave pulse, y = asin(kx – ωt) is incident on a rigid boundry, there is a phase difference of π between incident and reflected wave.
∴reflected wave is
y_r = a sin(kx – ωt + π)
= -a sin(kx – ωt)

(ii) When a wave pulse y = a sin(kx – ωt) is incident on an open boundry, there is no phase difference between incident and reflected wave.
y_r = a sin(kx – ωt + Φ)
Φ = 0,
y_r = a sin(kx – ωt)

Question 57.
When does a wave suffering reflection undergo a phase change of π?
Answer:
The reflected wave undergo a phase change of π when the reflection takes place at a rigid boundry.

Question 58.
When does a wave suffering reflection does not undergo any phase change?
Answer:
The reflected wave does not undergo any phase change when the reflection takes place at open boundry.

Question 59.
When are stationary waves produced?
Answer:
Stationary waves are produced due to superposition of two identical waves of same amplitude and frequencies travelling with the same speed but in opposite direction.

Question 60.
Explain the formation of stationary waves.
Answer:
Consider a wave travelling along positive x-axis given by y₁ = a sin(kx – ωt) when the wave get reflected, the reflected wave travels along negative x-direction and is given by y₂ = asin(kx + ωt)
When the two waves superpose, the resultant wave y is
y = y₁ + y₂
y = a sin(kx – ωt) + a sin(kx + ωt)
y = a {sin(kx – ωt) + sin(kx + ωt)}
But

y = 2a sin kx cosωt
y = (2a sin kx) cos ωt
In the equation 2a sin kx represents amplitude of the resultant wave.
The resultant amplitude is minimum when sin kx = 0
or kx = 0, π, 2π, 3π
kx = nπ where n = 0, 1, 2, 3,……
But k = \(\frac{2 \pi}{\lambda}\)
\(\frac{2 \pi}{\lambda}\) x = nπ
or \(x=\frac{n \lambda}{2}\)
Such points where the resultant amplitude is zero are called nodes.
The resultant amplitude is maximum when sin kx = ±1
or

Such points where the resultant amplitude is maximum are called antinodes.

Question 61.
What are nodes in a stationary wave?
Answer:
Nodes are the position of particles in a stationary wave which are always at rest i.e., amplide is zero.

Question 62.
What are antinodes in a stationary wave?
Answer:
Antinodes are the position of particles in a stationary wave which vibrates with maximum amplitude.

Question 63.
What is the distance of seperation between a node and its neighbouring antinode?
Answer:
The separation between a node and its neighbouring antinode is \(\frac{\lambda}{4}\), where λ is the wavelength of the component waves.

Question 64.
If λ is the wavelength of the component waves. What is the distance between any two consecutive nodes or antinodes?
Answer:
The distance between any two nodes or antinodes is λ/2.

Question 65.
In a longitudinal stationary wave, at what points the pressure change is (a) maximum (b) minimum.
Answer:
In a stationary wave
(a) the pressure change is maximum at nodes
(b) the pressure change is minimum at antinodes.

Question 66.
What is the state of vibration of the particle midway between two nodes?
Answer:
The state of vibration is maximum at a point midway between the nodes.

Question 67.
What is the state of vibration of the particle midway between two antinodes?
Answer:
The state of vibration is minimum i.e., zero midway between two antinodes.

Question 68.
Discuss the modes of vibrations of stretched a string.
Answer:
Consider a string of a length L fixed at two ends A and B. When the string is plucked, a wave travel towards fixed ends with a velocity v. The wave gets reflected at fixed ends. The superposition of incident and reflected ends leads to transverse stationary waves.
When the string vibrates with two nodes and an antinode i.e., one segment, it is said to be in its fundamental node or first harmonic.

Length of the string L = \(\frac{\lambda_{1}}{2}\)
⇒ λ₁ = 2L
∴fundamental frequency v₂ = \(\frac{v}{\lambda_{1}}=\frac{v}{2 L}\)
When the string vibrates with 3 nodes and 2 antinodes, i.e., it vibrates with two segments, it is said to be in its first overtone or second harmonic


When the string vibrates with 4 nodes and 3 antinodes, i.e., it vibrates with 3 segments, it is said to be in its second
overtone or third harmonic


In general when the string vibrates n segment, it is said to be in n th harmonic.
frequency v_n = \(\frac{n v}{2 L}\)

Question 69.
What are beats?
Answer:
The waxing and waning or periodic rise and fall in the intensity of sound due to superposition of two sound waves of nearly same frequency and travelling in the same direction is called beats.

Question 70.
Define the terms
(i) beat period
(ii) beat frequency.
Answer:
(i) The time interval between any two successive waxing or waning is called beat period.
(ii) The number of beats heard per second is called beat frequency.

Question 71.
How is beat frequency related to the frequency of the component frequencies?
Answer:
The beat frequency is equal to the difference in frequencies of component frequencies.
v_b = v₁ ∼ v₂
ω_b = ω₁ – ω₂
ω₁ = 2πv₁, ω₂ = 2πv₂ and ω_b = 2πv_b
∴2πv_b = 2πv₁ – 2πv₂
2πv_b =2π(v₁ – v₂)
v_b = v₁ – v₂

Question 72.
Show that the beat frequency is equal to the difference between the frequencies of the component waves.
Answer:
Consider two sound waves of same amplitude ‘a’ but nearly same angular frequencies ω1 and ω2.
The displacement produced by the two waves are
y₁ = a cosωt
and y₂=a cosω₂t
Using the principle of superposition, the resultant displacement is
y = y₁ + y₂
y = a cos ω₁t + a cos ω₂t

Question 73.
What property of waves is behind the phenomenon of beats?
Answer:
The phenomenon of superposition of waves is behind the phenomenon of beats.

Question 74.
What is the maximum beat frequency that the human ear can detect?
Answer:
The maximum beat frequency that a normal human ear can detect is 10 Hz.

Question 75.
What happens to frequency of the tuning fork if one of the prongs of a tuning fork is (a) filed (b) loaded?
Answer:
(a) If one of the prongs is filed the frequency increases
(b) If one of the prongs is loaded the frequency decreases.

Question 76.
Define the terms (i) fundamental frequency (ii) harmonics (iii) overtones.
Answer:
(i) lowest frequency emitted by a body is called fundamental frequency.
(ii) The frequencies which are integral multiples of the fundamental frequency are called harmonics.
(iii) frequencies greater than the fundamental frequency are called overtones.

Question 77.
What is an closed pipe? Discuss the modes of vibration of an column in an open pipe?
Answer:
A pipe which is closed at one end and open at the other end is called closed pipe.

When longitudinal waves are produced at the mouth of a closed pipe, the waves travel along the length of the pipe in the form of compressions and rarefactions and gets reflected at the closed end. The superposition of incident and reflected waves leads to longitudinal stationary waves.

Case (i): When the air column vibrates with only one node and one antinode, it is said to be in its fundamental mode or first harmonic. The corresponding frequency is called fundamental frequency V₁.

Case (ii) In the next mode of vibration, the air column vibrates with two nodes and two antinodes. The air column is said to be in its third harmonic or first over tone.

Case (iii) In the next mode of vibration, the air column vibrates with three nodes and antinodes. The air column is said to be in its fifth harmonic or second overtone.

∴the ratio of overtones are the odd harmonics of the fundamental.

Question 78.
What is an open pipe? Discuss the modes of vibration of air in an open pipe.

Answer:
Consider an open pipe of length L. When longitudinal wave is produced at one end of a pipe, the wave travels along the length of the pipe and gets reflected at the other end. The superposition of incident and reflected wave leads to stationary wave.
Case (i) When the air column vibrates with two antinodes and a node in between, it is said be in its fundamental mode. The corresponding frequency is called fundamental frequency.
The length of the pipe L = \(\frac{\lambda_{1}}{2}\) ⇒ λ₁ = 2L the fundamental frequency v = \(\frac{v}{\lambda_{1}}=\frac{v}{2 L}\) …….(1)

Case (ii): In the next mode of vibration the air column vibrates with three antinodes and two nodes, the air column is said to be in its second harmonic or first overtone.
The length of the pipe \(L=\frac{2 \lambda_{2}}{2} \Rightarrow \lambda_{2}=\frac{2 L}{2}\)
∴frequency \(v_{2}=\frac{v}{\lambda_{2}}=\frac{2 v}{2 L}=2 v_{1}\) …….(2)

Case (iii): In the next mode of vibration the air column vibrates with four antinodes and three nodes, the air column is said to be in third harmonic or second overtone.

From equation (1), (2) and (3)
v₁: v₂:v₃ ………..:: 1:2:3…….
∴ratio of overtones are harmonics of the fundamental.

Question 79.
What is Doppler effect?
Answer:
The apparent change in the frequency of sound due to relative motion, between the source and the observer is called Doppler effect.

Question 80.
Obtain an expression for apparent frequency when the source move towards a stationary observer.
Answer:
Consider a source S emitting sound waves of frequency v and wavelength λ. Let the velocity of sound waves be v.
then v = vλ
where v is called true frequency and λ is the true wavelength.
Suppose the source move towards the observer with a velocity vs. After one second, the source will be at S’ covering a distances – v_s. Now the same number of waves v will occupy a distance (v – v_s). Then the wavelength will decrease to λ’
\(\lambda^{\prime}=\frac{v-v_{s}}{v}\)
The apparent frequency v’ is
\(\mathrm{v}^{\prime}=\frac{v}{\lambda^{\prime}}=\left(\frac{v}{v-v_{s}}\right) V\)

Thus the apparent frequency increases when the source move towards stationary listener.

Question 81.
Obtain an expression for apparent frequency when the source the source move away from a stationary observer.
Answer:
Consider a source of sound S emitting sound waves of frequency v and wavelength λ. Let v be the velocity of sound.
then v = vλ
where v is called true frequency and λ is the true wavelength.
Suppose the source away from the observer with a velocity vs. After one second, the source will be at S’ covering a distance vs. Now the same number of waves v will occupy a distance v + vs. Then the wavelength will increase to λ’.
\(\lambda^{\prime}=\frac{v+v_{s}}{v}\)

The apparent frequency
\(v^{\prime}=\frac{v}{\lambda^{\prime}}=\left(\frac{v}{v+v_{s}}\right) v\)

Therefore the apparent frequency decreases when the source move away from a stationary observer.

Question 82.
Obtain an expression for apparent frequency when the observer is moving towards a stationary source.
Answer:
Consider a source of sound S emitting sound waves of frequency v and wavelength λ. Let V be the velocity of sound.
then v = vλ
where v is called true frequency and λ is the true wavelength.
Let the observer move towards stationary source with a velocity V₀. Now the observer receives more number of waves in one second.

∴the apparent frequency increases when the observer move towards stationary source.

Question 83.
Obtain an expression for apparent frequency when the observer move away from stationary source.
Answer:
Consider a source of sound S emitting sound waves of frequency v and wavelength λ. Let v be the velocity of sound.
then v = vλ
where v is called true frequency and λ is the wavelength.
Let the observer move away from the source with a velocity V₀. Now the observer receives less number of waves in one second.
apparent frequency \(v^{\prime}=v-\frac{v_{0}}{\lambda}\)

∴apparent frequency decreases when the observer move away from the stationary source.

Question 84.
Obtain an expression for apparent frequency when source and the observer in motion.
Answer:
Consider a source of sound S emitting sound waves of frequency v and wavelength λ. Let v be the velocity sound.
then v = vλ
where v is the true frequency and λ is the true wavelength.
Let the source and observer move with velocities v_s and v₀ along + x-axis.
In one second, the source travels a distance Vs and the wave travel a distance V. The number of waves emitted by source in one second is v.
These waves occupy a distance (V – V_s)

The wavelength due to the motion of source is λ’ = \(\frac{v-v_{\mathrm{s}}}{v}\)
If the observer is moving with a velocity V₀, the relative velocity and sound with respect to the moving observer is (v – v₀)
∴apparent frequency
v’ = \(\left(\frac{v-v_{0}}{\lambda^{\prime}}\right)\)
= \(\left(\frac{v-v_{0}}{v-v_{s}}\right) v\)

Question 85.
Mention few applications of Doppler effect.
Answer:

• Doppler effect is used in estimating the velocities of aeroplanes, submarines etc.,
• Used by traffic police to measure the speed of the vehicles.
• Used to estimate the speed of distance stars, planets and celestial bodies.
• Used to study heart beats and blood flow in different parts of the body.
• Used to get information about motion of blood and pulsation of heart valves.
• Used to get information about pulsation of the heart of the foetus.

Question 86.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one of the string. How long does the distances take to reach the other end?
Answer:
mass M = 2.50 kg
tension T = 200 N
Length L = 20.0 m
Velocity of transverse wave v = \(\sqrt{\frac{T}{\mu}}\)

∴time taken by the wave to reach the other end is 0.5 s.

Question 87.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top. Given the speed of sound in air is 340ms^-1 (g = 9.8 ms^-2)
Answer:
h = 300 m
v = 340 ms^-1
g = 9.8 ms^-2
Let t₁ is the time taken by the stone to reach the pond.
t₂ is the time taken by the sound to reach the top.
∴time after which sound is heard t = t₁ +t₂
to find t₁

\(x=v_{0} t+\frac{1}{2} a t^{2}\)
x = h = 300m, v0 =0, a = g = 9.8 ms^-2, t = t₁ = ?
300 = 0 + \(\frac{1}{2}\) × 9.8 t₁²
\(t_{1}^{2}=\frac{2 \times 300}{9.8}\)
t₁ = 7.825 s
to find t₂

Question 88.
A steel wire has a length 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 ms^-1.
Answer:
L = 12.0 m
M = 2.10 kg
v = 343 ms^-1
T = ?

Question 89.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound (b) the transmitted sound? Speed of sound in air is 340 ms^-1 and in water 1486 ms^-1.
Answer:
v = 1000 kHz = 1000 × 10³ Hz = 10 Hz.
(a) v = vλ

(b) v = vλ

Question 90.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 ms^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
v = 1.7 kms^-1 = 1.7 × 10 ms^-1
γ = 4.2 MHz = 4.2 × 10⁶ Hz
λ = ?
\(\lambda=\frac{v}{v}=\frac{1 \cdot 7 \times 10^{3}}{4 \cdot 2 \times 10^{6}}\)
= 0.4048 × 10^-3 m = 0.4048 mm

Question 91.
A transverse harmonic wave on a string is described by y (x, t) = 3.0 sin (36t + 0.018 x = π/4) where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or stationary wave? If it is travelling, what are the speed and direction of its propagation.
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two consecutive crests in the wave?
Answer:
The given equation y(x,t) = 3sin(36t + 0.018x + \(\frac{\pi}{4}\)) ……..(1)
is the standared form y(x,t) = Asin(ωt + kx + Φ) …….(2)
Hence it is a travelling wave. It is travelling from right to left.
Comparing equation (1) and (2)
A = 3 cm
ω = 36 rad s^-1
k = 0.018 rad cm^-1
Φ = \(\frac{\pi}{4}\) rad
(a) Velocity
\(v=\frac{\omega}{k}=\frac{36}{0.018}\)
v = 2000 cm s^-1
(b) amplitude A = 3 cm
ω = 2πv
frequency v = \(\frac{\omega}{2 \pi}=\frac{36}{2 \pi}=5.729 \mathrm{Hz}\)
(c) Initial phase at the origin Φ = \(\frac{\pi}{4}\) rad
(d) least distance between two consecutive crests in the wave
i.e., wavelength \(\lambda=\frac{V}{v}=\frac{2000}{5 \cdot 729}=349.1 \mathrm{cm} \mathrm{s}^{-1}\)
= 3.491 ms^-1

Question 92.
For the travelling harmonic wave y(x,t) = 2.0 cos 2π(10t – 0.0080 × 0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of (a) 4 m (b) 0.5 m (c) \(\frac{\lambda}{2}\) (d) \(\frac{3 \lambda}{4}\).
Answer:
The given equation for a travelling wave y(x,t) = 2 cos2π(10t – 0.0080 × 0.35)
= 2cos(20πt – 2π × 0.008x + 2π × 0.35) ………(1)
is in the standared form
y(x,t) = 2cos(ωt + kx + Φ)
comparing (1) and (2),
k = 2π x 0.008
\(\frac{2 \pi}{\lambda}\) = 2π x 0.008
(i) when x = 4m = 400 cm
phase diff Φ = \(\frac{2 \pi}{\lambda}\) × path difference
= 2π × 0.008 × 400 = 6.4π rad

(ii) when x = 0.5 m = 50 cm
Φ = \(\frac{2 \pi}{\lambda}\) × x = 2π × 0.008 × 50 = 0.8π rad

Question 93.
The transverse displacement of a string (clamped at its both ends) is given by
y(x,t) = 0 06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos(120πt) where x and y are in m and t in s. The length of the string is 15 m and its mass is 3 × 10^-2 kg.
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency and speed of each wave?
(c) Determine the tension in the string.
Answer:
(a) in the given equation y(x,t) = 0 06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos(120πt) involves the function x and t, it represents a stationary wave.
(b) The stationary wave is in standard form y(x,t) = 2 Asin kx cosωt
This stationary wave is due to super positions of two waves.
y₁(x,t) = Asin(kx – ωt) and y₂(x,t) = asin(kx + ωt) travelling in opposite direction.
k = \(\frac{2 \pi}{3}\)
\(\frac{2 \pi}{\lambda}=\frac{2 \pi}{3}\)
λ = 3 m
ω = 120 π
2πv = 120 π
v = 60 Hz

Question 94.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10^-2 kg and its linear mass density is 4.0^-2 × 10 kg m^-1. What is (a) the speed of a transverse wave on the string and (b) the tension in the string?
Answer:
(a) v = 45 Hz
M =3.5 × 10^-2 kg
µ = 4 × 10^-2 kg m^-1
v = \(\frac{1}{2 L} \sqrt{\frac{T}{\mu}}\)
But \(\sqrt{\frac{T}{\mu}}\) = v
v = \(\frac{v}{2 L}\)
⇒ v = 2Lv
\(\mu=\frac{M}{L} \Rightarrow L=\frac{M}{\mu}=\frac{3 \cdot 5 \times 10^{-2}}{4 \times 10^{-2}}=0.875 \mathrm{m}\)
∴speed v = 2Lv = 2 × 0.875 x 45
= 78.75 ms^-1
(b) v = \(\sqrt{\frac{T}{\mu}}\) ⇒ T = v²µ = 78.75² × 4 × 10^-2
T = 248 N

Question 95.
A metre – long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz). When the tube length 25.5 cm and 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.
Answer:
Since the open pipe shows resonance for 25.5 cm and 79.3 cm, these two position represents first and third harmonic
then 79.3 – 25.5 = \(\frac{\lambda}{4}\)
25.5 = \(\frac{\lambda}{4}\)
λ = 102 cm = 1.02 m
∴speed of sound v = vλ
v = 340 × 1.02 = 346.8 ms^-1

Question 96.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Since steel rod is clamped at its middle, a node is formed at the middle which the antinode is formed at the ends. Therefore the length of the rod

L = \(\frac{\lambda}{4}+\frac{\lambda}{4}=\frac{\lambda}{2}\)
⇒ \(\frac{\lambda}{2}\) = 100 cm
λ = 200 cm = 2 m
v = vλ = 2.53 × 10³ × 2
= 5.06 × 10³ = 5.06km s^-1

Question 97.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe of both ends are open (speed of sound in air is 340 ms^-1)
Answer:
In case of a pipe closed at one end, frequency in the n^th mode is
v_n =(2n – 1)\(\frac{v}{4 L} \)
(2n -1) = \(\frac{v_{n} \times 4 L}{v}\)
= \(\frac{430 \times 2 \times 0 \cdot 2}{340}\)
(2n – 1) = 1.01
2n = 2.02
n = 1
∴pipe vibrates in its first harmonic.

Question 98.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 ms^-1 (b) recedes from the platform with a speed of 10 ms^-1 (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms^-1.
Answer:
(a) v = 400 Hz, v_s = 10 ms^-1, v = 340 ms^-1
apparent frequency
v’ = \(\left(\frac{v}{v-v_{s}}\right) v\)
= \(\left(\frac{340}{340-10}\right) 400\)
= 412.1 Hz

Question 99.
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz. What is the frequency of B?
Answer:
v_b =v_A ± v_B
b_B = v_A ± v_B
= 324 ± 6
= 318 Hz or 330 Hz
∴When the tension of A is reduced, the number of beats reduces to 2. This happens only when frequency of B is 318 Hz.
∴v_B = 318 Hz

Question 100.
Use the formula v = \(\sqrt{\frac{\gamma P}{\rho}}\) to explain why the speed of sound in air (a) is independent of pressure (b) increases with temperature (c) increases with humidity.
Answer:
v = \(\sqrt{\frac{\gamma P}{\rho}}\)
density ρ = \(\frac{M}{V}\)
∴ v = \(\sqrt{\frac{\gamma P V}{M}}\)
When T is PV = constant
∴velocity of sound is independent of pressure.

(b) v = \(\sqrt{\frac{\gamma P}{\rho}}\)
But PV = RT => P = \(\frac{R T}{V}\)
\(v=\sqrt{\frac{\gamma R T}{\rho V}}=\sqrt{\frac{\gamma R T}{M}}\)
∴v ∝ \(\sqrt{T}\)
∴velocity sound increases with temperature.

(c) humidity is the amount of water vapour present in air.
∴Velocity of sound in moist air
\(v_{m}=\sqrt{\frac{\gamma P}{\rho_{m}}}\) ……(1)
velocity of the sound in dry air
\(v_{d}=\sqrt{\frac{\gamma P}{\rho d}}\)
\(\frac{v_{m}}{v_{d}}=\sqrt{\frac{\rho d}{\rho m}}\)
∵density of water vapour is less than dry air
∴ρ_m < ρ_d
Hence v_m > v_d.
∴velocity of sound increases with humidity.

Question 101.
A train standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to be station with a speed of 10 ms^-1. What are the frequency, wavelength and the speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the ease when the air is still and the observer runs towards the yard at a speed of 10 ms^-1. The speed of sound in still air can be taken as 340 ms^-1.
Answer:
v = 400 Hz
v_w= 10 ms^-1
v₀ – 10 ms^-1
v = 340 ms^-1
For an observer standing on the platform.
frequency v = 400 Hz
wave length λ = \(\frac{v+v_{w}}{\gamma}=\frac{340+10}{400}=\frac{350}{400}\)
= 0.875 m
speed of sound =v + v_w= 340 +10 = 350 ms^-1
When the air is still and the observer run towards the yard at a speed 10 ms^-1
frequency
v’ = \(\left(\frac{v+v_{L}}{v}\right) v\)
= \(\left(\frac{340+10}{340}\right) 400\)
= 411.8 Hz
wave length λ = \(\frac{v}{r^{\prime}}=\frac{340}{411 \cdot 8}\)
velocity of the sound v = 340 ms^-1

(b) v = 400 Hz, v_s = + 10 ms^-1, V = 340 ms^-1

(ii) The speed of sound in each case is 340 ms^-1.