In an organ pipe whose one end is at x = 0,th | vedclass

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(C) The pressure variation in a standing wave is given by $p(x, t) = p_0 \cos(kx) \sin(\omega t)$.At a closed end,the displacement is zero,which corre

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closed at both ends,length $= 2 \ m$

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(C) The pressure variation in a standing wave is given by $p(x, t) = p_0 \cos(kx) \sin(\omega t)$.At a closed end,the displacement is zero,which corresponds to a pressure antinode (maximum pressure amplitude).At an open end,the pressure is atmospheric,which corresponds to a pressure node (zero pressure amplitude).Given $p = p_0 \cos \frac{3\pi x}{2} \sin 300\pi t$.At $x = 0$,$p = p_0 \cos(0) \sin(300\pi t) = p_0 \sin(300\pi t)$. Since the pressure amplitude is maximum $(p_0)$,$x = 0$ is a closed end.For the other end to be a closed end,the pressure amplitude must be maximum,i.e.,$|\cos \frac{3\pi x}{2}| = 1$.This occurs when $\frac{3\pi x}{2} = n\pi$ for integer $n$.For the first possible length $L > 0$,we set $n = 3$,so $\frac{3\pi L}{2} = 3\pi$,which gives $L = 2 \ m$.Thus,the pipe is closed at both ends with a length of $2 \ m$.

In an organ pipe whose one end is at $x = 0$,the pressure is expressed by $p = p_0 \cos \frac{3\pi x}{2} \sin 300\pi t$,where $x$ is in meters and $t$ is in seconds. The organ pipe can be:

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