Question

A closed organ pipe and an open 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

${\nu }_c=\frac{v}{4L_c}$

where $v$ is the velocity of sound in air and $L_c$ is the length of the close pipe.

For an open organ pipe, the frequency of fundamental mode is

${\nu }_o=\frac{v}{2L_o}$

where $L_o$ is the length of the open pipe.

Given, $L_c=L_o$

$\therefore {\nu }_o=2{\nu }_c$

and ${\nu }_o-{\nu }_c=2$

Solving the above equations we get:-

${\nu }_o=4Hz$

${\nu }_c=2Hz$

When the length of the open pipe is halved, its frequency of fundamental mode is

${\nu }_o^{\prime }=\frac{v}{2(\frac{L_o}{2})}=2{\nu }_o=2\times 4Hz=8Hz$

When the length of the closed pipe is doubled, its frequency of fundamental mode is

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

Hence, Number of beats produced per second $={\nu }_o^{\prime }-{\nu }_c^{\prime }=8-1=7$