question_answer

Two sources A and B are sending notes of frequency 680 Hz. A listener moves from A and B with a constant velocity u. If the speed of sound in air is \[340\text{ }m{{s}^{-1}}\], what must be the value of us that he hears 10 beats per second?

A)  \[2.0m{{s}^{-1}}\]

B)  \[2.5m{{s}^{-1}}\]

C)  \[3.0m{{s}^{-1}}\]

D)  \[3.5m{{s}^{-1}}\]  

Correct Answer: B

Solution :

Listener go from \[A\to B\] with velocity (u) let the apparent frequency of sound from source A by listener \[n=n\left( \frac{v-{{v}_{0}}}{v+{{v}_{s}}} \right)\] or \[n=680\left( \frac{340-u}{340+0} \right)\] The apparent frequency of sound from source B by listener \[n=n\left( \frac{v+{{v}_{0}}}{v-{{v}_{s}}} \right)=680\left( \frac{340+u}{340-0} \right)\] But listener hear 10 beats per second. Hence,         \[n-n=10\] or \[680\left( \frac{340+u}{340} \right)-680\left( \frac{340-u}{340} \right)=10\] or \[2(340+u-340+u)=10\] or \[u=2.5m{{s}^{-1}}\]