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(C) In the fundamental mode of a closed pipe,the length of the pipe $L$ corresponds to one-fourth of the wavelength,i.e.,$L = \frac{\lambda}{4}$.Given

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$\frac{1}{4 t}$

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(C) In the fundamental mode of a closed pipe,the length of the pipe $L$ corresponds to one-fourth of the wavelength,i.e.,$L = \frac{\lambda}{4}$.Given that the time taken for the sound wave to travel the length $L$ is $t$,we have $t = \frac{L}{v}$,where $v$ is the speed of sound.Since $L = \frac{\lambda}{4}$,we can write $t = \frac{\lambda}{4v}$.The time period $T$ of the wave is given by $T = \frac{\lambda}{v}$.Substituting $\lambda = vT$ into the expression for $t$,we get $t = \frac{vT}{4v} = \frac{T}{4}$.Therefore,the time period $T = 4t$.The frequency $f$ is the reciprocal of the time period,so $f = \frac{1}{T} = \frac{1}{4t}$.

In the fundamental mode,the time required for a sound wave to reach the closed end of an air-filled pipe is $t$ seconds. What is the frequency of vibration of the air column?

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