A) \[\frac{v+u}{v-u}\]
B) \[\frac{2u}{v+u}\]
C) \[\frac{2v}{v+u}\]
D) \[\frac{u+v}{2v}\]
Correct Answer: B
Solution :
Let the frequency of sound sent by the source\[=f\] So, the frequency of sound observed by a person near the wall will be \[f=\left( \frac{v-(-u)}{v-0} \right){{f}_{0}}=\frac{v+u}{v}{{f}_{0}}\] The frequency of sound reflected by the wall \[f\,=\frac{v-0}{v-u}f\] \[\Rightarrow \] \[f\,=\frac{v}{v-u}\times \frac{v+u}{v}f\] \[\Rightarrow \] \[f\,=\frac{v+u}{v-u}f\] Initial wavelength\[{{\lambda }_{i}}=\frac{v}{f}\] Final wavelength\[{{\lambda }_{f}}=\frac{v}{f\,}=\frac{v}{f}\left( \frac{v-u}{v+u} \right)\] So, fractional change in the wavelength is \[\frac{\Delta \lambda }{{{\lambda }_{i}}}=\frac{\frac{v}{f}\left[ \frac{v-u}{v+u}-1 \right]}{\frac{v}{f}}=\frac{2u}{v+u}\]
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