Wave Motion Ques 61
- A closed organ pipe has a fundamental frequency of $1.5 kHz$. 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 $20,000 Hz$ )
(a) 7
(b) 4
(c) 5
(d) 6
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Answer:
Correct Answer: 61.(d)
Solution:
Formula:
- Fundamental frequency of closed organ pipe is given by $f _0=v / 4 L$, where $v$ is the velocity of sound in it and $L$ is the length of the pipe.
Also, overtone frequencies are given by
$$ f=(2 n+1) \frac{v}{4 L} \quad \text { or } \quad f=(2 n+1) f _0 $$
Given that, $f _0=1500 Hz$ and $f _{\max }=20000 Hz$
This means, $f _{\max }>f$
So, $\quad f _{\max }>(2 n+1) f _0$
$\Rightarrow \quad 20000>(2 n+1) 1500$
$\Rightarrow \quad 2 n+1<13.33 \Rightarrow 2 n<13.33-1$
$\Rightarrow \quad 2 n<12.33$ or $n<6.16$
or $n=6$ (integer number)
Hence, total six overtones will be heard.
BETA
