(C) The fundamental frequency of an open pipe is given by $f_1 = \frac{v}{2l}$.Given: $v = 330 \ m/s$,$l = 33 \ cm = 0.33 \ m$.$f_1 = \frac{330}{2 \ti

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(C) The fundamental frequency of an open pipe is given by $f_1 = \frac{v}{2l}$.Given: $v = 330 \ m/s$,$l = 33 \ cm = 0.33 \ m$.$f_1 = \frac{330}{2 \times 0.33} = \frac{330}{0.66} = 500 \ Hz$.The given frequency is $f = 1000 \ Hz$.We know that $f = p \times f_1$,where $p$ is the harmonic number.$1000 = p \times 500 \implies p = 2$.Thus,the frequency corresponds to the second harmonic of the pipe.

Class 11 Physics Waves Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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An open pipe of length $33 \ cm$ resonates with a frequency of $1000 \ Hz$. If the speed of sound is $330 \ m/s$,then this frequency is:

A
Fundamental frequency of the pipe
B
Third harmonic of the pipe
C
Second harmonic of the pipe
D
Fourth harmonic of the pipe

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Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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An open pipe of length $33 \ cm$ resonates with a frequency of $1000 \ Hz$. If the speed of sound is $330 \ m/s$,then this frequency is:

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What is the minimum length of a tube,open at both ends,that resonates with a tuning fork of frequency $350 \ Hz$? [Velocity of sound in air $= 350 \ m/s$] ..... $cm$

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