Question

A closed organ pipe and an open organ pipe of same length produce $2$ beats/second while vibrating in their fundamental modes. The length of the open organ pipe is halved and that of closed pipe is doubled. Then, the number of beats produced per second while vibration in the fundamental mode is

• A. 2
• B. 6
• C. 8
• D. 7

Answer 1

Answer: (D)

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

$\because L_C=L+O$ (Given)

$\therefore {\nu }_o=2{\nu }_c$ ......(i)

${\nu }_o-{\nu }_c=2$ (Given) ......(ii)

Solving (i) and(ii), we get

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

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

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

Hence, member of beats produced per second

$={\nu }_o^{\prime }-{\nu }_c^{\prime }=8-1=7$