Question

A closed organ pipe has a fundamental frequency of $1.5\text{kHz}$. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is $20,000\text{Hz}$)

• A. $6$
• B. $4$
• C. $7$
• D. $5$

Answer 1

The correct option is A $6$

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 $20,000=(2N-1)\times 1500$
$\Rightarrow N=7.1\approx 7$
$\therefore$Number of over tones = (mode of vibration)$-1=7-1=6$.