QJEEM23Phd66a: JEE Main 2023 Physics - Class 11 JEEM23Phd66a Solution: An organ pipe 40   cm long is open at both ends. The speed …

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 $900 \, \text{Hz}$ .