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(D) In a closed pipe of length $L$, the fundamental mode (first harmonic) has a node at the closed end and an antinode at the open end.For the fundame

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$\frac{1}{4t}$

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(D) In a closed pipe of length $L$, the fundamental mode (first harmonic) has a node at the closed end and an antinode at the open end.For the fundamental mode, the length of the pipe is $L = \frac{\lambda}{4}$, where $\lambda$ is the wavelength.The time $t$ taken for the sound wave to travel from the open end to the closed end (a distance $L$) is given by $t = \frac{L}{v}$, where $v$ is the speed of sound.Since $v = f\lambda$, we have $t = \frac{L}{f\lambda}$.Substituting $L = \frac{\lambda}{4}$, we get $t = \frac{\lambda/4}{f\lambda} = \frac{1}{4f}$.Rearranging for frequency $f$, we get $f = \frac{1}{4t}$.

In the fundamental mode, the time required for a sound wave to reach the closed end of a pipe filled with air is $t$ seconds. What is the frequency of vibration of the air column? ($\lambda$ = wavelength of the wave)

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