Waves

JEE Main level, numerical answer, medium

91 of 142

Question 91 of 142

An organ pipe 40cm40cm long is open at both ends. The speed of sound in air is 360ms−1360m{s}^{-1}. The frequency of the second harmonic is _________Hz.

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Answer: 900

The formula for the frequency in an open organ pipe is
f=nv2Lf=\dfrac{nv}{2L}
Since, it is a second harmonic, n=2n=2
The frequency of the second harmonic becomes f=2v2L=vLf=\dfrac{2v}{2L}=\dfrac{v}{L}
The data given is
L=0.4mv=360ms−1L=0.4m v=360m{s}^{-1}
Substituting the values in the frequency formula
f=360ms−10.4m=900Hzf=\dfrac{360m{s}^{-1}}{0.4m}=900Hz

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