An observer moves towards a stationary source of sound with a speed 1/5th of the speed of sound. The wavelength and frequency of the source emitted are and f respectively. The apparent frequency and wavelength recorded by the observer are respectively
- 1.2f, 1.2
- 1.2f,
- f, 1.2
- 0.8f, 0.8
Answer
(2) 1.2f,
Solution
Observer moving towards a stationary source with vo = v5:
f' = f ( v + vov )
= f ( v + v/5v )
= 1.2f
∵ The source is at rest, the waves in the medium are not compressed or stretched; the observer only meets them faster.
∴ The wavelength stays .
Answer: (2)