Question

A fork of frequency 256 Hz resonates with a closed organ pipe of length 25.4 cm. If the length of pipe be increases by 2 mm, the number of beats/sec will be :

• A. 4
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

Frequency of the fundamental mode of a closed organ pipe is:
${\nu }_{c0}=\frac{v}{4l}=256$
So, $v=256\times 4\times 25.4$ cm/s

Now with the increased length the frequency becomes
${\nu }_{c0}^{\prime }=\frac{v}{4l^{\prime }}=\frac{256\times 4\times 25.4}{4\times 25.6}=254$

Number of beats $={\nu }_{c0}-{\nu }_{c0}^{\prime }=2$