A) \[f,1.2\lambda \]
B) \[0.8f,0.8\lambda \]
C) \[1.2f,1.2\lambda \]
D) \[1.2f,\lambda \]
Correct Answer: D
Solution :
When an observer moves towards an stationary source of sound, then apparent frequency heard by the observer increases. The apparent frequency heard in this situation \[f'=\left( \frac{v+{{v}_{o}}}{v-{{v}_{s}}} \right)f\] As source is stationary hence,\[{{v}_{s}}=0\] \[f'=\left( \frac{v+{{v}_{o}}}{v} \right)f\] Given, \[{{v}_{o}}=\frac{v}{5}\] Substituting in the relation for\[f',\]we have \[f'=\left( \frac{v+v/5}{v} \right)f=\frac{6}{5}f=1.2f\] Motion of observer does not affect the wavelength reaching the observer, hence, wavelength remains\[\lambda \]. Note: When the speed of source and observer are much lesser than that of sound, the change in frequency becomes independent of the fact whether the source is moving or the observer.
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