Question

A closed organ pipe and an organ pipe of same length produce 2 beats/second when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced will be

• A. $7$
• B. $4$
• C. $8$
• D. $2$

Answer 1

The correct option is A $7$

For a closed organ pipe, the frequency of fundamental mode is $f_c=\frac{v}{4L_c}$ where ${}^{\prime }{v}^{\prime }$ is velocity of sound in air and $L_c$ is length of the closed pipe.

For an open pipe, the frequency of fundamental mode is

$f_0=\frac{v}{2L_0}$ Where ${}^{\prime }{L}_0^{\prime }$ is length of open pipe

$\because L_c=L_0$ (given)

$f_0=2f_c$............(1)

$f_0-f_c=2$...........(2)

Solving (1) and (2) we get $f_0=4Hg;f_c=2Hg$

when length of open pipe is halved, its frequency of fundamental mode is

$f_0^{\prime }=\frac{v}{2\left(\frac{L_0}{2}\right)}=2f_0=2\times 4Hg=8Hg$

When Length of closed pipe is doubled, its frequency

$f_c^{\prime }=\frac{v}{4\left(2L_c\right)}=\frac{1}{2}{f}_c=\frac{1}{2}\times 2Hz=1Hz$

Number of beats produced per second

$=f_0^{\prime }-f_c^{\prime }=8-1=7$