question_answer

A closed organ pipe of length 20 cm is sounded with tuning fork in resonance. What is the frequency of tuning fork? (y = 332 m/s)

A) 300 Hz                                  

B) 350 Hz

C) 375 Hz                                  

D) 415 Hz

Correct Answer: D

Solution :

Key Idea: When length of air column is \[\frac{\lambda }{4},\] then first resonance occurs. 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 air-column increases very much. This is the state of resonance. At first resonance     \[l=\frac{\lambda }{4}\] So, frequency of tuning fork                            \[f=\frac{v}{\lambda }=\frac{v}{4l}\] Given, I = 20 cm\[l=20\,cm=0.2\,m,\,v=332\,m/s\] Hence, \[f=\frac{332}{4\times 0.2}\,=415\,Hz\]