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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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