Waves and Sound MCQs with Answers and Detailed Explanations

In a transverse wave, particles of the medium move at right angles to the direction in which the wave travels.

For example, when a rope is shaken up and down, the disturbance moves forward while rope particles move vertically.

The wave carries energy, but the medium particles only oscillate around their equilibrium positions.

In a longitudinal wave, particles vibrate back and forth in the same direction as the wave travels.

Sound in air is a common example. It contains regions of compression and rarefaction.

Compression is where particles are closer together, while rarefaction is where particles are farther apart.

Frequency and period are reciprocal quantities.

\[
T=\frac{1}{f}
\]

Here,

\[
f=25\text{ Hz}
\]

Therefore,

\[
T=\frac{1}{25}
\]

\[
T=0.04\text{ s}
\]

A frequency of 25 Hz means the wave completes 25 oscillations every second.

Wave speed is given by:

\[
v=f\lambda
\]

Substitute the values:

\[
v=15\times2.0
\]

\[
v=30\text{ m/s}
\]

The speed depends on frequency and wavelength. In a fixed medium, changing frequency also changes wavelength so that wave speed remains determined by the medium.

A wave transports energy through a medium, but it does not carry each particle of the medium forward with it.

For a transverse wave on a rope, each marked point moves up and down around its equilibrium position.

The disturbance moves along the rope, while the rope material stays near its original average position.

Amplitude is the maximum displacement of particles from equilibrium.

For many mechanical waves, the energy carried is proportional to the square of amplitude:

\[
E\propto A^2
\]

Therefore, a larger amplitude means more energy transfer.

However, wave speed is mainly determined by the medium, not by amplitude for ordinary linear waves.

One complete wavelength corresponds to a phase difference of:

\[
360^\circ
\]

Therefore, a separation of \(\dfrac{\lambda}{4}\) gives:

\[
\Delta\phi=
\frac{\lambda/4}{\lambda}\times360^\circ
\]

\[
\Delta\phi=90^\circ
\]

Points separated by \(\dfrac{\lambda}{2}\) are \(180^\circ\) out of phase.

A travelling wave can be written as:

\[
y=A\sin(\omega t-kx)
\]

Comparing with the given equation:

\[
\omega=4\pi
\]

Angular frequency and frequency are related by:

\[
\omega=2\pi f
\]

Therefore:

\[
f=\frac{\omega}{2\pi}
\]

\[
f=\frac{4\pi}{2\pi}
\]

\[
f=2\text{ Hz}
\]

The wave completes two full oscillations each second.

At a fixed end, the string cannot move vertically.

Therefore, the reflected pulse must have opposite displacement so that the total displacement at the fixed point remains zero.

An upward pulse reflects as a downward pulse.

This is called phase inversion during reflection from a fixed boundary.

The superposition principle states that the resultant displacement is the algebraic sum of individual wave displacements.

If two upward displacements meet, constructive interference occurs.

If an upward displacement meets an equal downward displacement, destructive interference occurs.

After overlapping, the waves continue moving according to their original motion in a linear medium.

Sound is a mechanical wave.

It travels by making particles of a medium vibrate and pass energy to neighboring particles.

A vacuum has no particles, so there is no material available to transfer the vibration.

Therefore, sound cannot travel through outer space, although electromagnetic waves such as light can.

A travelling wave has the standard form:

\[
y=A\sin(\omega t-kx)
\]

The wave speed is:

\[
v=\frac{\omega}{k}
\]

From the equation:

\[
\omega=8\pi\text{ rad/s}
\]

and:

\[
k=2\pi\text{ rad/m}
\]

Therefore:

\[
v=\frac{8\pi}{2\pi}
\]

\[
v=4\text{ m/s}
\]

The sign \(-kx\) shows that the wave travels in the positive \(x\)-direction.

One wavelength corresponds to a phase difference of:

\[
360^\circ
\]

Therefore:

\[
\Delta\phi=
\frac{3\lambda/8}{\lambda}\times360^\circ
\]

\[
\Delta\phi=
\frac{3}{8}\times360^\circ
\]

\[
\Delta\phi=135^\circ
\]

Phase difference tells how far two particles are apart in their oscillation cycle.

For a string fixed at both ends, the fundamental mode has:

\[
L=\frac{\lambda_1}{2}
\]

Therefore:

\[
\lambda_1=2L
\]

\[
\lambda_1=2(1.20)
\]

\[
\lambda_1=2.40\text{ m}
\]

Frequency is:

\[
f_1=\frac{v}{\lambda_1}
\]

\[
f_1=\frac{240}{2.40}
\]

\[
f_1=100\text{ Hz}
\]

The fundamental frequency is the lowest possible standing-wave frequency.

For a string fixed at both ends, the \(n\)-th harmonic has:

\[
n
\]

antinodes and:

\[
n+1
\]

nodes, including the two fixed ends.

Therefore, the third harmonic has:

\[
3
\]

antinodes.

It also has four nodes: one at each end and two inside the string.

An open pipe has displacement antinodes at both ends.

For the fundamental mode:

\[
L=\frac{\lambda_1}{2}
\]

Therefore:

\[
\lambda_1=2L
\]

\[
\lambda_1=2(0.85)
\]

\[
\lambda_1=1.70\text{ m}
\]

Now:

\[
f_1=\frac{v}{\lambda_1}
\]

\[
f_1=\frac{340}{1.70}
\]

\[
f_1=200\text{ Hz}
\]

An open pipe supports all harmonics: \(f_1, 2f_1, 3f_1,\dots\).

A closed pipe has a node at the closed end and an antinode at the open end.

It supports only odd harmonics:

\[
f_1,\ 3f_1,\ 5f_1,\dots
\]

The next allowed frequency is the third harmonic:

\[
f_3=3f_1
\]

\[
f_3=3(85)
\]

\[
f_3=255\text{ Hz}
\]

The second harmonic does not occur in an ideal closed pipe.

Wave speed on a stretched string is:

\[
v=\sqrt{\frac{T}{\mu}}
\]

If tension changes from \(T\) to \(4T\), then:

\[
v’=\sqrt{\frac{4T}{\mu}}
\]

\[
v’=2\sqrt{\frac{T}{\mu}}
\]

\[
v’=2v
\]

Wave speed is proportional to the square root of tension, not directly proportional to tension.

A phase difference of:

\[
180^\circ
\]

means one wave has maximum positive displacement when the other has equal maximum negative displacement.

Therefore, their displacements cancel.

\[
A_{\text{resultant}}=A+(-A)
\]

\[
A_{\text{resultant}}=0
\]

This is complete destructive interference.

The energy is not destroyed; it is redistributed to other regions where constructive interference occurs.

Beat frequency is the absolute difference between two nearby frequencies.

\[
f_{\text{beat}}=\left|f_1-f_2\right|
\]

Therefore:

\[
f_{\text{beat}}=\left|256-262\right|
\]

\[
f_{\text{beat}}=6\text{ Hz}
\]

The listener hears six rises and falls in loudness every second.

Sound level difference is:

\[
\Delta\beta=10\log_{10}\left(\frac{I_2}{I_1}\right)
\]

Here:

\[
\frac{I_2}{I_1}=100
\]

Therefore:

\[
\Delta\beta=10\log_{10}(100)
\]

Since:

\[
\log_{10}(100)=2
\]

we get:

\[
\Delta\beta=20\text{ dB}
\]

A 20 dB increase means intensity becomes 100 times larger, although perceived loudness does not increase by exactly 100 times.

For a moving source approaching a stationary observer, the wavelength in front of the source becomes shorter.

The Doppler formula is:

\[
f’=
f\left(\frac{v}{v-v_s}\right)
\]

Here:

\[
f=600\text{ Hz}
\]

\[
v=340\text{ m/s}
\]

\[
v_s=20\text{ m/s}
\]

Therefore:

\[
f’=
600\left(\frac{340}{340-20}\right)
\]

\[
f’=
\frac{600\times340}{320}
\]

\[
f’=637.5\text{ Hz}
\]

The observed frequency is higher because the approaching source compresses the wavefronts.

For a moving observer approaching a stationary source:

\[
f’=
f\left(\frac{v+v_o}{v}\right)
\]

Here:

\[
f=500\text{ Hz}
\]

\[
v=340\text{ m/s}
\]

\[
v_o=34\text{ m/s}
\]

Therefore:

\[
f’=
500\left(\frac{340+34}{340}\right)
\]

\[
f’=
500\left(\frac{374}{340}\right)
\]

\[
f’=550\text{ Hz}
\]

The observer encounters more wavefronts every second, so the heard frequency increases.

The observer is behind the source, so the source moves away from the observer.

However, the observer moves toward the wavefronts from behind at exactly the same speed as the source.

Using the Doppler formula:

\[
f’=
f\left(\frac{v+v_o}{v+v_s}\right)
\]

Since:

\[
v_o=v_s
\]

we get:

\[
f’=
f\left(\frac{v+v_s}{v+v_s}\right)
\]

\[
f’=f
\]

The source stretching of wavelength is exactly balanced by the observer’s increased rate of meeting wavefronts.

This is a double Doppler effect.

First, the moving car acts as an observer approaching the stationary source:

\[
f_1=
400\left(\frac{340+20}{340}\right)
\]

\[
f_1=
400\left(\frac{360}{340}\right)
\]

The car then behaves like a moving source approaching the original source. The reflected frequency received by the original source is:

\[
f_2=
f_1\left(\frac{340}{340-20}\right)
\]

Therefore:

\[
f_2=
400\left(\frac{360}{340}\right)
\left(\frac{340}{320}\right)
\]

\[
f_2=
400\left(\frac{360}{320}\right)
\]

\[
f_2=450\text{ Hz}
\]

The reflected sound has a higher frequency because the car moves toward the source during both Doppler shifts.

Sound speed in still air is measured relative to the air.

When air moves relative to the ground, sound is carried with the moving medium.

Therefore:

\[
v_{\text{ground}}=
v_{\text{sound relative to air}}+v_{\text{wind}}
\]

\[
v_{\text{ground}}=340+20
\]

\[
v_{\text{ground}}=360\text{ m/s}
\]

The frequency produced by a stationary source does not automatically change because of wind. Wind changes the propagation speed relative to the ground.

For a string fixed at both ends, the \(n\)-th harmonic has:

\[
n+1
\]

total nodes, including the two end nodes.

For the fifth harmonic:

\[
\text{total nodes}=5+1=6
\]

Two nodes are at the fixed ends.

Therefore, internal nodes are:

\[
6-2=4
\]

The fifth harmonic also has five antinodes.

Frequency is determined by the source and must remain continuous at the boundary.

Wave speed on a string is:

\[
v=\sqrt{\frac{T}{\mu}}
\]

For the second string:

\[
v_2=
\sqrt{\frac{T}{4\mu}}
\]

\[
v_2=\frac{v_1}{2}
\]

Since:

\[
v=f\lambda
\]

and frequency remains unchanged, the wavelength becomes half in the heavier string.

The amplitude may also change because some wave energy is reflected at the junction.

For an open pipe of length \(L\):

\[
f_{\text{open}}=\frac{v}{2L}
\]

For a closed pipe of the same length:

\[
f_{\text{closed}}=\frac{v}{4L}
\]

Therefore:

\[
f_{\text{closed}}=
\frac{1}{2}f_{\text{open}}
\]

Thus:

\[
f_{\text{closed}}=\frac{f}{2}
\]

A closed pipe has a node at its closed end and an antinode at its open end.

For two coherent waves with equal intensity:

\[
I_{\text{resultant}}
=
I_1+I_2+2\sqrt{I_1I_2}\cos\phi
\]

Since:

\[
I_1=I_2=I
\]

and:

\[
\phi=60^\circ
\]

we get:

\[
I_{\text{resultant}}
=
I+I+2I\cos60^\circ
\]

\[
I_{\text{resultant}}
=
2I+2I\left(\frac{1}{2}\right)
\]

\[
I_{\text{resultant}}=3I
\]

The result is between complete constructive interference, which gives \(4I\), and complete destructive interference, which gives zero intensity.

For a sinusoidal wave on a string, average power is proportional to amplitude squared:

\[
P_{\text{avg}}\propto A^2
\]

If amplitude changes from \(A\) to \(2A\), then:

\[
P’_{\text{avg}}\propto(2A)^2
\]

\[
P’_{\text{avg}}\propto4A^2
\]

Therefore:

\[
P’_{\text{avg}}=4P_{\text{avg}}
\]

Doubling amplitude does not merely double the energy transfer rate. It increases average power by a factor of four.