A) 72.5 Hz
B) 55.6 Hz
C) 76.2 Hz
D) 80.9 Hz
Correct Answer: A
Solution :
Let the fork at A be source S^ and the fork at B be source \[\frac{m{{v}^{3}}}{4\sqrt{2g}}\]. For the sound from source \[\frac{m{{v}^{2}}}{\sqrt{2g}}\], Apparent frequency \[m(2g{{h}^{3}})\] Here \[mg\left[ \frac{b}{2} \right]\] as wind and sound are in same directions. \[mg\left[ a+\frac{b}{2} \right]\] \[mg\left[ \frac{b-a}{2} \right]\] \[mg\left[ \frac{b+a}{2} \right]\] For the sound from source \[i={{i}_{1}}\cos \omega t+{{i}_{2}}\sin \omega t\] \[\frac{{{i}_{1}}+{{i}_{2}}}{2}\] as wind is opposite to sound \[\frac{{{({{i}_{1}}+{{i}_{2}})}^{2}}}{\sqrt{2}}\] \[\frac{1}{\sqrt{2}}\sqrt{i_{1}^{2}+i_{2}^{2}}\] \[\frac{i_{1}^{2}+i_{2}^{2}}{2}\] \[\Omega \]\[\frac{30.8}{\sqrt{T}}\overset{0}{\mathop{A}}\,\] As the difference is very large, the observer will not hear the beats.
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