A) 436 to 586
B) 426 to 574
C) 426 to 584
D) 436 to 674
Correct Answer: A
Solution :
The linear velocity of Whistle \[{{v}_{S}}=r\omega =1.2\times 2\pi \frac{400}{60}=50\,m/s\] When Whistle approaches the listener, heard frequency will be maximum and when listener recedes away, heard frequency will be minimum So, \[{{n}_{\max }}=n\,\left( \frac{v}{v-{{v}_{s}}} \right)=500\,\left( \frac{340}{290} \right)=586Hz\] \[{{n}_{\min }}=\,n\,\left( \frac{v}{v+{{v}_{s}}} \right)=500\,\left( \frac{340}{390} \right)=436Hz\]
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