First slide

Doppler

Question

An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency ?

Moderate

A

zero
B

0.5%
C

5%
D

20%

Solution

When observer moves towards stationary source, then the apparent frequency is given by
$n^{'} = n (\frac{v + v_{O}}{v}) or \frac{n^{'}}{n} = (\frac{v + v_{O}}{v}) Here v_{o} = \frac{V}{5} = \frac{320}{5} = 64 m / s \frac{n^{'}}{n} = (\frac{320 + 64}{320}) = \frac{384}{320} e r e f o r e, p e r c e n t a g e i n c r c a s e (\frac{n^{'} - n}{n}) = (\frac{n^{'}}{n} - 1) = (\frac{384}{320} - 1) \times 100 \begin{aligned} = (\frac{384 - 320}{320}) \times 100 = (\frac{64}{320} \times 100) \\ = 20 % \end{aligned}$

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