JEE MAIN - Physics (2023 - 8th April Morning Shift - No. 20)
An organ pipe $$40 \mathrm{~cm}$$ long is open at both ends. The speed of sound in air is $$360 \mathrm{~ms}^{-1}$$. The frequency of the second harmonic is ___________ $$\mathrm{Hz}$$.
Answer
900
Explanation
An organ pipe that is open at both ends resonates at all harmonics, including the fundamental (first harmonic), second harmonic, third harmonic, etc.
The frequency $f$ of the $n$-th harmonic for a pipe open at both ends is given by:
$f_n = \frac{n v}{2L}$,
where:
- $n$ is the number of the harmonic,
- $v$ is the speed of sound, and
- $L$ is the length of the pipe.
To find the frequency of the second harmonic ($n = 2$), we can substitute the given values into the formula:
$f_2 = \frac{2 \times 360}{2 \times 0.4} = 900 \, \text{Hz}$.
Therefore, the frequency of the second harmonic is $900 \, \text{Hz}$.
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