Question

An observer moves towards a stationary source of sound with a velocity one tenth the velocity of sound.The apparent increase in frequency is

Answer 1

Answer on Question #41140, Physics, Mechanics | Kinematics | Dynamics

Question:

An observer moves towards a stationary source of sound with a velocity one tenth the velocity of sound. The apparent increase in frequency is

Answer:

The Doppler effect (or Doppler shift is the change in frequency of a wave for an observer moving relative to its source. In classical physics the relationship between observed frequency $f$ and emitted frequency $f_0$ is given by:

$$f = \frac{c + v_r}{c + v_s} f_0$$

where $c$ is the velocity of waves in the medium;

$v_r$ is the velocity of the receiver relative to the medium; positive if the receiver is moving towards the source (and negative in the other direction);

$v_s$ is the velocity of the source relative to the medium; positive if the source is moving away from the receiver (and negative in the other direction).

In this case $v_r = 0.1\,c$, $v_s = 0$ therefore

$$f = \frac{c + 0.1\,c}{c} f_0 = 1.1\,f_0$$

Increase in frequency equals:

$$\Delta f = f - f_0 = 0.1\,f_0$$

Answer: $0.1\,f_0$

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