Question

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 meter and $t$ in sec. The organ pipe can be

• A. closed at one end, open at another with length $0.5m$
• B. open at both ends, length $=1m$
• C. closed at both ends, length $=2m$
• D. closed at one end, open at another with length $=\frac{2}{3}m$

Answer 1

Answer: (C)

closed at both ends, length $=2m$

$p=p_0\cos (3\pi \ast X/2)\sin (300\pi \ast t)$

at $X=0.5m$

$p=p_{0}cos\left(3\pi /4\right)\sin (300\pi \ast t)$ (neither maxima nor minima so neither

close nor open)

at $X=1.0m$

$p=p_{0}cos\left(3\pi /2\right)\sin (300\pi \ast t)=0$ (minima so seems to be a possible close)

at $X=2.0m$

$p=p_{0}cos\left(3\pi \right)\sin (300\pi \ast t)=-p_0\sin (300\pi \ast t)$ (max pressure means

seems to be a possible close)

which is true.

at $X=0m$

$p=p_{0}cos\left(0\right)\sin (300\pi \ast t)=p_0\sin (300\pi \ast t)$ (close at one end )

at $X=2/3$

$p=p_0\cos \left(\pi \right)\sin (300\pi \ast t)=-p_0\sin (300\pi \ast t)$ (close at other end too)

so option "D" is incorrect,
only "C" is correct.