Question

A tube of length $1\text{m}$ containing $H_2$ is closed at one end. If the velocity of sound in air be $340\text{m/s}$. Then, the fundamental frequency and the next higher overtone in $H_2$ are:

• A. $85\text{Hz},170\text{Hz}$
• B. $170\text{Hz},255\text{Hz}$
• C. $85\text{Hz},255\text{Hz}$
• D. $55\text{Hz},255\text{Hz}$

Answer 1

The correct option is C $85\text{Hz},255\text{Hz}$

For a closed organ pipe only odd harmonic will exist.
The fundamental frequency is,
$f_1=\frac{v}{4L}=\frac{340}{4\times 1}=85\text{Hz}$
The next higher overtone will be $3^{\text{rd}}$ harmonic or $1^{\text{st}}$ overtone.
$f_3=\frac{3v}{4L}=3f_1=3\times 85=255\text{Hz}$
$\therefore \left(c\right)$ is the correct option.