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(A) In a closed organ pipe of length $L$,the fundamental mode corresponds to a quarter wavelength,i.e.,$L = \lambda / 4$,which implies $\lambda = 4L$.

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$(4t)^{-1}$

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(A) In a closed organ pipe of length $L$,the fundamental mode corresponds to a quarter wavelength,i.e.,$L = \lambda / 4$,which implies $\lambda = 4L$.The time $t$ taken for the sound wave to travel from the open end to the closed end is the time taken to cover the distance $L$ at the speed of sound $v$. Thus,$t = L / v$,or $L = vt$.Substituting $L = vt$ into the wavelength expression,we get $\lambda = 4vt$.The frequency $f$ of the vibration is given by $f = v / \lambda$.Substituting the value of $\lambda$,we get $f = v / (4vt) = 1 / (4t) = (4t)^{-1}$ Hz.Therefore,the correct option is $A$.

In the fundamental mode,the time required for a sound wave to travel from the open end to the closed end of an air-filled pipe is $t$ seconds. What is the frequency of vibration of the air column? (Neglect end correction)

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