A) \[f,1.2\lambda \]
B) \[0.8f,0.8\lambda \]
C) \[1.2f,1.2\lambda \]
D) \[1,f2\,\lambda \]
Correct Answer: A
Solution :
When an observer moves towards an stationary source of sound then apparent frequency heard by the observer increases. The the apparent frequency heard in this situation \[f=\left( \frac{v+{{v}_{0}}}{v-{{v}_{s}}} \right)f\] As source is stationary hence\[{{V}_{s}}=0\] \[f=\left( \frac{v+{{v}_{0}}}{v} \right)f\] given \[{{v}_{0}}=\frac{v}{5}\] \[f=\left( \frac{v+v/5}{v} \right)f=\frac{6}{5}=1.2f\] Substituting in the relation for f we have \[f=\left( \frac{v+v/5}{v} \right)f=\frac{6}{5}=1.2f\] Motion of observer does not. affect the wavelength reaching the observer hence wavelength remains\[\lambda \].
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