### Question

Source S produces sound waves of speed \(v\) and frequency \(f\). S moves with constant velocity \(\frac{v}{5}\) away from a stationary observer.

What is the frequency measured by the observer?

A. \(\frac{4}{5} f\)

B. \(\frac{5}{6} f\)

C. \(\frac{6}{5} f\)

D. \(\frac{5}{4} f\)

**▶️Answer/Explanation**

Ans:B

The frequency observed by a stationary observer when a source is in motion can be calculated using the Doppler effect formula for sound:

\[f’ = \frac{f \cdot (v + v_o)}{v + v_s},\]

- \(f\) is the frequency emitted by the source.
- \(v\) is the speed of sound.
- \(v_o\) is the velocity of the observer, which is stationary (0).
- \(v_s\) is the velocity of the source, which is moving away at \(\frac{v}{5}\).

Now, substitute these values into the formula:

\[f’ = \frac{f \cdot (v + 0)}{v + \frac{v}{5}}.\]

Simplify the equation:

\[f’ = \frac{f \cdot v}{\frac{6}{5}v} = \frac{5}{6}f.\]).

*Question*

*v*away from the gun.

*f*and speed

*c*. Which of the following is the frequency of the microwaves measured at the gun after reflection by the car?

**Answer/Explanation**

### Markscheme

**change in frequency in term of gun is**

\(\Delta f’=-\frac{v}{c}{f_o}\)

\(f’-f_o=-\frac{v}{c}{f_o}\)

\(f’=f_o -\frac{v}{c}{f_o}\)

Frequency observed by Gun after reflection from car

\(\frac{\Delta f”}{f’}=-\frac{v}{c}\)

or

\(\Delta f” =-\frac{v}{c}{f’} =-\frac{v}{c} \times (f_o -\frac{v}{c}{f_o})\)

\(f”-f’ =f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)

\(f” =f’+f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)

\(=f_o -\frac{v}{c}{f_o} +f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)

\(=f_o(1-2\frac{v}{c} +\frac{v^2}{c^2})\)

A train moving at speed *u* relative to the ground, sounds a whistle of constant frequency *f* as it moves towards a vertical cliff face.

The sound from the whistle reaches the cliff face and is reflected back to the train. The speed of sound in stationary air is *c*.

What whistle frequency is observed on the train after the reflection?

A. \(\frac{{(c + u)}}{{(c – u)}}f\)

B. (c + u)*f*

C. (c – u)*f*

D. \(\frac{{(c – u)}}{{(c + u)}}f\)

**Answer/Explanation**

Answer – A

frequency recieved by cliff face:

\(f{}’=f\times \frac{c}{c-\mu }\)

then after reflection ,cliff face acts as source,whole train act as listener

Frequency received by train :

\(f{}”=f{}’\times \frac{c+\mu }{c }\)

\(f{}”=f{}’\times \frac{c }{c -\mu }\times \frac{c+\mu }{c}\)

\(f{}”=f\times \frac{c+\mu }{c-\mu }\)

Light of wavelength *λ*_{0} is emitted from a nearby galaxy. The light is received on Earth and the wavelength is measured to be *λ* where *λ*<*λ*_{0}. Which of the following correctly describes the speed and direction of motion of the galaxy?

**Answer/Explanation**

## Markscheme

A

## Examiners report

The vast majority of candidates understood that the fractional change in the wavelength was needed and hence discounted B or D. But it would seem that they did not read the stem carefully, where it is clear that the wavelength has decreased – indicating that it is travelling towards Earth.

### Question

A standing wave is formed between two loudspeakers that emit sound waves of frequency f.

A student walking between the two loudspeakers finds that the distance between two consecutive sound maxima is \(1.5 \mathrm{~m}\). The speed of sound is \(300 \mathrm{~m} \mathrm{~s}^{-1}\).

What is \(f\) ?

A. \(400 \mathrm{~Hz}\)

B. \(200 \mathrm{~Hz}\)

C. \(100 \mathrm{~Hz}\)

D. \(50 \mathrm{~Hz}\)

**▶️Answer/Explanation**

Ans:C

In a standing wave, the distance between consecutive maxima (or minima) is half of the wavelength (\(\lambda/2\)). Given that the distance between two consecutive sound maxima is \(1.5 \, \text{m}\), we can find the wavelength (\(\lambda\)):

\(\lambda/2 = 1.5 \, \text{m}\)

\(\lambda = 2 \cdot 1.5 \, \text{m} = 3 \, \text{m}\)

Now, we can use the formula for the speed of a wave:

\[v = f \cdot \lambda\]

Let’s solve for \(f\):

\[300 \, \text{m/s} = f \cdot 3 \, \text{m}\]

Now, divide both sides by 3:

\[f = \frac{300 \, \text{m/s}}{3 \, \text{m}} = 100 \, \text{Hz}\]

So, the frequency (\(f\)) of the sound waves is \(100 \, \text{Hz}\).