Question

An open pipe vibrating in its fundamental frequency is suddenly closed at one end. As a result, the frequency of the third harmonic of the closed pipe is found to be higher by $100\text{Hz}$. The fundamental frequency of the open pipe is :-

• A. $200\text{Hz}$
• B. $30\text{Hz}$
• C. $240\text{Hz}$
• D. $480\text{Hz}$

Answer 1

The correct option is A $200\text{Hz}$

Modes of vibration of an air column in a closed organ pipe are given by
$f_{nc}=(2n+1)\frac{v}{4L}$ where $n=1,2,\mathrm{3...}$
The third harmonic of closed pipe is given by
$f_{3c}=\frac{3v}{4L}$
The modes of vibration of an open organ pipe are given by
$f_{no}=\frac{nv}{2L}$ where $n=1,2,\mathrm{3...}$
Fundamental frequency of open organ pipe is
$f_{1o}=\frac{v}{2L}$
According to the question,
$f_{3c}-f_{1o}=100$
$\Rightarrow \frac{3v}{4L}-\frac{v}{2L}=100$
$\Rightarrow \frac{v}{4L}=100\Rightarrow \frac{v}{L}=400$

Hence, the fundamental frequency of open organ pipe is
$f_{1o}=\frac{v}{2L}=\frac{1}{2}\left(\frac{v}{L}\right)=\frac{1}{2}\times 400=200\text{Hz}$