WBCHSE Class 11 Physics Doppler Effect In Sound and Light Question and Answers

Doppler Effect In Sound and Light  Question and Answers

Question 1. Each of the two men A and B is carrying a source of sound of frequency. If A approaches B with a velocity u,

1. How many beats per second will be heard by A and
2. How many beats per second will be heard by B? (Velocityofsound =c)

Answer:

Apparent frequency, n’ = \(\frac{c+u_o}{c-u_s} \times n\)

The distance between A and B is decreasing. So, the velocity of the listener u₀ and that of the source u_s, are both positive.

1. When A is the listener, velocity of the listener, u₀ = u; velocity of the source B, u_s = 0

So, \(n_A=\frac{c+u}{c} \times n\)

Evidently, n_A > n

∴ Number of beats per second

= \(n_A-n=\left(\frac{c+u}{c}-1\right) \times n=\frac{u}{c} n .\)

2. When B is the listener, velocity of the listener, u₀ = 0; velocity of the source A, u_s = u

So, \(n_B=\frac{c}{c-u} \times n\)

Evidently, \(n_B>n\)

∴ Number of beats per second

= \(n_B-n=\left(\frac{c}{c-u}-1\right) \times n\)=\(\frac{u}{c-u} n\)

Question 2. A car Is approaching a hill at a high speed. At that time, if the horn of the car is blown, the driver hears the echo sharper than the original sound. Explain the reason.
Answer:

Answer: The original sound after being reflected from the hill approaches the car and the driver listens to the echo. So, in this case, the listener is moving towards the source of the echo. So, due to the Doppler effect, the echo is of a higher apparent frequency. Thus, it appears to be sharper than the original sound to the driver.

Question 3. Certain characteristic wavelength in the light from a galaxy has a longer wavelength compared to that from a terrestrial source. Is the galaxy approaching or receding?
Answer:

The galaxy is receding. It can be concluded from the increase in wavelength, i.e., decrease in frequency due to the Doppler effect, that the distance between the source of light (the galaxy) and the observer (the earth) is gradually increasing. Hence, the galaxy is receding.

Question 4. Show that the apparent frequency f’ of a source of sound moving with a speed vs towards a stationary receiver is \(f^{\prime}=\frac{f c}{c-v_s}\), where c is the velocity of sound and f is the frequency.
Answer:

If the source of sound approaches the stationary listener with velocity v_s, the number of sound waves f produced per second occupies the distance c – v_s.

∴ Apparent wavelength, \(\lambda^{\prime}=\frac{c-v_s}{f}\)

So, apparent frequency, f’ = \(\frac{c}{\lambda^{\prime}}=\frac{f c}{c-v_s}\)

Question 5. Two sources, each emitting a sound of wavelength A, are kept at a fixed distance. How many beats will be heard by a listener moving with a velocity u along the line joining the two sources?
Answer:

If n is the frequency and v is the velocity of sound, the apparent frequency to a listener in motion for a stationary source,

n’ = \(\frac{v+u}{v} \times n\)

The velocities of the listener for the two sources in question are -u and +u.

So, \(n_1^{\prime}=\frac{v-u}{v} \times n \text { and } n_2^{\prime}=\frac{v+u}{v} \times n\)

∴ Number of beats per second

= \(n_2^i-n_1^{\prime}=\frac{2 u}{v} \times n=\frac{2 u}{\frac{v}{n}}=\frac{2 u}{\lambda} .\)

Example 6. A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has a frequency of 2f. If v is the velocity of sound, what will be the velocity of the car?
Answer:

If u_s and u₀ are the velocities of the source of sound and the listener respectively,

the frequency of the echo, f’ = \(\frac{v+u_o}{v-u_s} \times f\)

Here, u_s = u₀ = u(say) and f’ = 2f

∴ 2f = \(\frac{v+u}{v-u} \times f \text { or, } v+u=2(v-u) \text { or, } 3 u=v \text { or, } u=\frac{v}{3} \text {. }\)

Example 7. What should be the velocity of a source of sound so that the apparent frequency to a listener will be half the actual frequency of the source? The velocity of sound in air = v.
Answer:

If n is the actual frequency, apparent frequency = \(\frac{n}{2}\).

Here the listener is stationary, i.e., u₀ = 0

Again decrease of frequency means that the source is receding from the listener. So the velocity of the source us is negative.

Therefore, from the relation n’ = \(\frac{v+u_o}{v-u_s} \times n\), we have

⇒ \(\frac{n}{2}=\frac{v}{v+u_s} \times n \text { or, } v+u_s=2 v \text { or, } u_s=v\)

i.e., the source is receding from the listener with the velocity of sound.

Question 8. What should be the velocity of a source of sound so that the apparent frequency to a listener will be twice the actual frequency of the source? The velocity of sound in air = v.
Answer:

Apparent Frequency, \(n^{\prime}=\frac{v+u_o}{v-u_s} \times n\)

In this case n’ = 2n. The listener is stationary; so u₀ = 0.

The apparent frequency is higher, i.e., the source is approaching the listener. So, the velocity of the source u_s is positive. Therefore,

2n = \(\frac{v}{\nu-u_s} \times n \text { or, } v=2\left(\nu-u_s\right) \text { or, } u_s=\frac{v}{2}\)

i.e., the source is approaching the listener with half the velocity of sound.

Question 9. What should be the velocity of a listener so that the apparent frequency of the sound coming from a stationary source to him will be twice the actual frequency? The velocity of sound in air = v.
Answer:

Apparent frequency, \(n^{\prime}=\frac{v+u_o}{v-u_s} \times n\)

In this case velocity of the source, u_s= 0.

Since the apparent frequency is higher, the listener is approaching the source, i.e., the velocity of the listener u₀ is positive.

Again, n’ = 2n.

∴ 2n = \(\frac{v+u_o}{v} \times n \text { or, } 2 v=v+u_o \text { or, } u_o=v\)

i.e., the listener is approaching the stationary source with the velocity of sound.

Question 10. Doppler effect gives an idea of a continuously expanding universe—explain.
Answer:

Light coming from the distant star can be analysed with the spectrometer. The experiment shows that the wavelength of a spectral line for a light source situated for away from Earth is greater than that for the same light source on Earth i.e., frequency is comparatively low.

From this apparent decrease in frequency, known as the redshift in the Doppler effect, we conclude that all distant stars are receding from Earth which indicates that the Universe is continuously expanding.

Question 11. A listener moving with constant velocity passes a stationary source. Draw a graph to show the change of apparent frequency of the source to the listener with time. The actual frequency of the source is n.
Answer:

While approaching the stationary source apparent frequency will be,

n’ = \(\frac{V+u_0}{V} n\)…(1)

[velocity of sound in air = V, velocity of listener = u₀]

While receding the stationary source apparent frequency will be,

n” = \(\frac{V-u_0}{V} n\)…(2)

from equations (1) and (2) it is clear that nf and n” remain constant with time and also n’ > n> n”. The change of apparent frequency with time is shown below.

Here OS denotes the time taken by the listener to pass by the source. After that, the listener continually moves away from the source.

Question 12. Both of a sound source and a listener are approaching each other with the same speed \(\frac{c}{10}\) (speed of sound in air =c). What will be the percentage of apparent increase or decrease in frequency of sound?
Answer:

Apparent frequency,

n’ = \(\frac{c+u_0}{c-u_S} n\)

So, the apparent increase in frequency,

n’ – n = \(\frac{u_0+u_S}{c-u_S} n\)

∴ Percentage of change in frequency

= \(\frac{n^{\prime}-n}{n} \times 100=\frac{u_0+u_S}{c-u_S} \times 100\)

= \(\frac{\frac{c}{10}+\frac{c}{10}}{c-\frac{c}{100}} \times 100=\frac{2}{9} \times 100=22.2 \%\)

Question 13. A band of music at a frequency f is moving towards a wall at a speed v_b. A motorist is following the band with a speed v_m. If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist.
Answer:

Two separate sounds will be heard by the motorist, one is direct from the band and the other is the echo from the wall.

Apparent frequency of direct sound, \(f_1=f\left(\frac{v+v_m}{v+v_b}\right)\)

Apparent frequency of the echo, \(f_2=f\left(\frac{v+v_m}{v-v_b}\right)\)

Therefore, the beat frequency heard by the motorist,

n = \(f_2-f_1=f\left(v+v_m\right) \cdot\left[\frac{1}{v-v_b}-\frac{1}{v+v_b}\right]\)

= \(\frac{2 f v_b\left(v+v_m\right)}{v^2-v_b^2}\)

Question 14. Why Doppler effect is clearly realised in the case of sound but not in the case of light waves?
Answer:

Our auditory system is more sensitive to realise a small change in audible frequency. On the other hand, our visual system is not so sensitive to detect such a small change in the visible frequency of light. In our daily life, the magnitude of our relative velocity of us with a light source on earth is not high enough to observe such a noticeable change in frequency, i.e., the Doppler effect.