The Doppler Effect - Example 3

Page 1

Moving source and moving observer

A bat emits ultrasonic waves with a frequency of 56 kHz, and the bat is travelling at 15m/s. If the bat is travelling towards a moth that is flying away at 5m/s, what frequency would the moth observe? Take the speed of sound in air to be 340 m/s.

Solution

Vo = 5m/s - use the minus sign because it's moving away from the source - smaller apparent frequency

Vs = 15 m/s - use minus because it's moving towards the observer - will create a bigger apparent frequency

f'=\(\frac{v±vo}{v±vs}\)f

f'=\(\frac{340-5}{340-15}\)56 = 57.7 kHz

Example 3