A closed organ pipe of length 1.2 , m vibrate | vedclass

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(B) For a closed organ pipe,the length $L$ is related to the wavelength $\lambda$ for the $n^{th}$ overtone as $L = (2n + 1) \frac{\lambda}{4}$.For th

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$0.4 \, m$ from the open end

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(B) For a closed organ pipe,the length $L$ is related to the wavelength $\lambda$ for the $n^{th}$ overtone as $L = (2n + 1) \frac{\lambda}{4}$.For the first overtone,$n = 1$,so $L = \frac{3\lambda}{4}$.Given $L = 1.2 \, m$,we have $\frac{3\lambda}{4} = 1.2 \, m$,which gives $\lambda = 1.6 \, m$.In a standing wave,pressure variation is maximum at the displacement nodes.In a closed organ pipe,the open end is a displacement antinode (pressure node),and the closed end is a displacement node (pressure antinode).The displacement nodes occur at distances $x = 0, \frac{\lambda}{2}, \lambda, \dots$ from the closed end.Alternatively,measuring from the open end,the displacement nodes occur at $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \dots$ from the open end.For the first overtone,the first displacement node (maximum pressure variation) is at $\frac{\lambda}{4} = \frac{1.6}{4} = 0.4 \, m$ from the open end.

$A$ closed organ pipe of length $1.2 \, m$ vibrates in its first overtone mode. The pressure variation is maximum at:

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