MCQ
An organ pipe is closed at one end has fundamental frequency of $1500 Hz$. The maximum number of overtones generated by this pipe which a normal person can hear is :
- A$14$
- B$13$
- ✓$6$
- D$9$
If a closed pipe vibration in ${N^{th}}$ mode then frequency of vibration
$n = \frac{{(2N - 1)v}}{{4l}} = (2N - 1){n_1}$ (where ${n_1} = $ fundamental frequency of vibration)
Hence $20,000 = (2N - 1) \times 1500$ ==> $N = 7.1 \approx 7$
Also, in closed pipe Number of over tones = (No. of mode of vibration $-1$)
$= 7 -1 = 6.$
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