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 ±e the state of resonance. At first resonance \[l=\frac{\lambda }{4}\] So, frequency of tuning fork \[f=\frac{v}{\lambda }=\frac{v}{4l}\] Given, \[l=20\,cm\,=0.2\,m\] \[v=332\,m/s\] Hence, \[f=\frac{0.332}{4\times 0.2}=415\,Hz\]You need to login to perform this action.
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