Question 290: Waves

Question

A source and listener are both moving towards each other with speed $\frac{\mathrm{V}}{10}$. (where V is speed of sound) If the frequency of sound note emitted by the source is ' $n$ ', then the frequency heard by the listener would be nearly

Accepted Answer

The frequency heard by a listener when both the source and the listener are moving towards each other can be calculated using the Doppler effect formula. When both the source and listener are in motion, the observed frequency $ f' $ is given by:
$$ f' = f \left( \frac{v + v_L}{v - v_S} \right) $$
where:
$ f $ is the emitted frequency,
$ v $ is the speed of sound,
$ v_L $ is the speed of the listener,
$ v_S $ is the speed of the source.
In this scenario, both the source and the listener are moving towards each other at a speed of $ \frac{V}{10} $.
So, substitute these values into the formula:
$$ v_L = \frac{V}{10}, \quad v_S = \frac{V}{10} $$
The observed frequency $ f' $ will be:
$$ f' = n \left( \frac{V + \frac{V}{10}}{V - \frac{V}{10}} \right) $$
Simplify the expression:
Numerator: $ V + \frac{V}{10} = \frac{10V}{10} + \frac{V}{10} = \frac{11V}{10} $
Denominator: $ V - \frac{V}{10} = \frac{10V}{10} - \frac{V}{10} = \frac{9V}{10} $
Thus, the expression becomes:
$$ f' = n \left( \frac{\frac{11V}{10}}{\frac{9V}{10}} \right) $$
Simplify further:
$$ f' = n \left( \frac{11}{9} \right) $$
Calculate the approximate value:
$$ \frac{11}{9} \approx 1.22 $$
Therefore, the frequency heard by the listener is approximately $ 1.22n $.
Answer: Option B: 1.22 n

Answer

1.1 n

Answer

1.22 n

Answer

n

Answer

1.27 n

Waves

A source and listener are both moving towards each other with speed \frac{\mathrm{V}}{10}. (where V is speed of sound) If the frequency of sound note emitted by the source is ' n ', then the frequency heard by the listener would be nearly

Choose the correct answer: