1.

A closes organ pipe of length 20 cm is sounded with a tuning fork in resonance. What is the frequency of the tuning fork? (v = 332 m/s)(a) 300 Hz (b) 350 Hz(c) 375 Hz (d) 415 Hz

Answer»

(d) 415 Hz 

So,

In resonance, the frequency of the fork is equal to the frequency of the organ pipe,

\(f = \frac{v}{4L}\) = \(\frac{332}{4 \times 0.2}\) = 415 Hz



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