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

Answer: (A)

7

Answer 2

Given, $f_{0}-f_{c}=2$...(i)
Frequency of fundamental mode for a closed organ pipe, $f_{c}=\frac{v}{4 L_{c}}$
Similarly frequency of fundamental mode for an open organ pipe, $f_{o}=\frac{v}{2 L_{o}}$
Given $L_{c}=L_{o}$
$\Rightarrow f_{0}=2 f_{c}$...(ii)
From Eqs. (i) and (ii), we get
$f_{o}=4\, Hz$ and $f_{c}= 2\, Hz$
When the length of the open pipe is halved, its frequency of fundamental mode is
$f_{0}'=\frac{v}{2\left[\frac{L_{o}}{2}\right]}=2 f_{o}=2 \times 4\, Hz =8\, Hz$
When the length of the closed pipe is doubled, its frequency of fundamental mode is
$f_{0}'=\frac{v}{4\left(\underline{2 L}_{c}\right)}=\frac{1}{2} f_{c}=\frac{1}{2} \times 2=1\, Hz$
Hence, number of beats produced per second is
$f_{0}'-f_{c}'=8-1=7$