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