question_answer

                Two sources are at a finite distance apart. They emit sounds of wavelength \[\lambda \]. An observer situated between them on line joining approaches one source with speed \[u\]. Then the number of beats heard/s by observer will be:                      

A)                                                                                                                                                         \[\frac{2u}{\lambda }\]

B)                 \[\frac{u}{\lambda }\]                   

C)                 \[\frac{u}{2\lambda }\]                

D)                 \[\frac{\lambda }{u}\]

Correct Answer: A

Solution :

                                Let \[v\] be the speed of sound and \[n\] the original frequency of each source.                 They emit sounds of wavelength \[\lambda \]                 When observer moves towards one source (say A), the apparent frequency of A as observed by the observer will be                 \[n'=n\left( \frac{v+u}{v} \right)\]                 The observes is now receding source B, so die apparent frequency of S observed will be                 \[n'=n\,\left( \frac{v-u}{v} \right)\]                 Thus, number of beats                 \[x=n'-n''\]                 \[=n\left[ \frac{v+u}{v}-\frac{v-u}{v} \right]\]                 \[=\frac{n}{v}[v+u-v+u]\]                 \[=\frac{2nu}{v}\] but         \[v=n\lambda \] \[Thus,x=\frac{2nu}{n\lambda }=\frac{2u}{\lambda }\]