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
Answer/Explanation
Markscheme
\(\Delta f’=-\frac{v}{c}{f_o}\)
\(f’-f_o=-\frac{v}{c}{f_o}\)
\(f’=f_o -\frac{v}{c}{f_o}\)
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 λ0 is emitted from a nearby galaxy. The light is received on Earth and the wavelength is measured to be λ where λ<λ0. 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 \(100 \, \text{Hz}\).