Question

The frequency of a closed pipe is $f_1$ if it vibrates with two nodes and is $f_2$ if it vibrates with three nodes. If the difference between the frequencies $f_1$ and $f_2$ is 200 Hz, then the length of the pipe is (Speed of sound in air $=340\,ms^{-1}$)

• A. $75\text{ cm}$
• B. $65\text{ cm}$
• C. $85\text{ cm}$
• D. $55\text{ cm}$

Hint:

For a closed pipe: \[ f=\frac{nv}{4L} \] where \(n=1,3,5,\ldots\). Only odd harmonics are present.

Answer 1

The Correct Option is B

Concept: For a closed organ pipe, only odd harmonics are produced. The frequencies are given by \[ f_n=\frac{nv}{4L} \] where \[ n=1,3,5,7,\ldots \] and \(L\) is the length of the pipe.

Step 1: Identify the harmonics corresponding to the given nodes.
For a closed pipe:
• Two nodes correspond to the third harmonic.
• Three nodes correspond to the fifth harmonic. Therefore, \[ f_1=\frac{3v}{4L} \] and \[ f_2=\frac{5v}{4L} \]

Step 2: Use the given frequency difference.
Given, \[ f_2-f_1=200 \] Substituting, \[ \frac{5v}{4L}-\frac{3v}{4L}=200 \] \[ \frac{2v}{4L}=200 \] \[ \frac{v}{2L}=200 \]

Step 3: Calculate the length of the pipe.
Using \[ v=340\,ms^{-1} \] \[ \frac{340}{2L}=200 \] \[ 340=400L \] \[ L=0.85\,m \] \[ L=85\,cm \] Hence, \[ \boxed{85\,cm} \]