question_answer

A source of sound is travelling at                                                                                                  \[\frac{100}{3}m{{s}^{-1}}\]  along a road, towards a point A. When the source is 2 m away from A, a person standing at a point 0 on a road perpendicular of frequency v'. The distance of 0 from A at that time is 4m. If the original frequency is 640 Hz, then the value of v' is  (Given that velocity of sound                                                                                                  \[=340\text{ }m{{s}^{-1}}\] )

A)  620 Hz                 

B)  680 Hz

C)  720 Hz                 

D)  840 Hz

Correct Answer: B

Solution :

                 Effective value of velocity of source,                                                                                        \[{{v}_{s}}=\frac{100}{3}\cos \theta\]                                                                         \[=\frac{100}{3}\times \frac{3}{5}=20\,m{{s}^{-1}}\]                                                                                                  \[v'=\frac{v}{v-{{v}_{s}}}v\]                                                                             \[v'=\frac{340}{340-20}\times 640\,Hz=680\,Hz\]