A closed organ pipe of length 20 cm is sounded with
tuning fork in resonance. What is the frequency of
tuning fork? `(v = 332 ms^(-1))`

Options

A. 300 Hz
B. 350 Hz
C. 375 Hz
D. 415 Hz

Answer

d

Explanation

If we adjust the length of air-column in closed organ
pipe as such its any natural frequency equals to the
frequency of tuning fork, then the amplitude of forced
vibrations of ais-column increases very much. This is the
state of resonance. `l = lambda/4`

So, frequency of tuning fork, `f=v/lambda = 4/(4l)`
Given, `l = 20 cm = 0.2 m and v = 332 ms^(-1)`
Hence, `f=(332)/(4xx 0.2) = 415` Hz