Question

An organ of length $L$ is open at one end closed at other end. The air column in the pipe is vibrating in second overtone. The minimum distance from the open end where the pressure amplitude is half of the maximum value is

• A. $L/4$
• B. $L/8$
• C. $L/12$
• D. $L/15$

Answer 1

The correct option is A $L/4$

$\lambda +\lambda /4=L$

$\frac{5\lambda }{4}=L$

at any position $x$ the presence in given by

$\Delta P=\Delta P_0\cos x\cos wt$

Here amplitude $A=\Delta P\cos kx=\Delta P_0\cos \frac{2\pi }{\lambda }x$

amplitude at distance $x$

$a=A\cos \frac{2\pi }{\lambda }x$

$a=\left(\frac{A}{1}\right)=A\cos \frac{2\pi }{4L/5}\times x$

$2\pi \times \left(\frac{5}{4L}\right)x={\cos }^{-1}\left(1/2\right)$

$2\pi \times \left(\frac{5}{4L}\right)x=\frac{\pi }{3}$

$x=\frac{2L}{15}$

For closed angle pipe (open at me end & closed at then)

$f=\frac{(2n-1)}{4}\left(\frac{v}{1}\right)$

for $2^{nd}$ from $n=3$

$f=\frac{(2\times 3-1)}{4}\left(\frac{V}{L}\right)=\frac{5}{4}\frac{V}{L}....\left(i\right)$

amplitude at any distance $x$

$a=A\cos \frac{2\pi }{\lambda }x$

$\lambda =\left(\frac{V}{f}\right)=\frac{V}{5V/2L}=\left(\frac{4L}{5}\right)$

$a=\left(\frac{A}{2}\right)=A\cos \frac{2\pi }{\left(4L/5\right)}x$

$\frac{\pi }{3}=\frac{2a\times 5}{24L}\times x$

$x=\frac{2L}{15}$