Question

A cylindrical tube of length one metre is divided into two parts by a thin flexible diaphragm in the middle. The ends of the tube are closed by a similar diaphragms. One portion of the cylinder is filled with hydrogen while the other portion is filled with oxygen. The two diaphragms are set in vibration of same frequency such that the middle diaphragm is motionless. Velocity of sound in hydrogen is 1100 m/sec and velocity of sound in oxygen is 300 m/sec. What should be the minimum frequency of the end diaphragms under these conditions?

• A. 1650 Hz
• B. 1100 Hz
• C. 2200 Hz
• D. 825 Hz

Answer 1

The correct option is A 1650 Hz

When diaphragms A and B are set in oscillation, antinodes are formed at A and B while a node is formed at C.
The portions AC and BC behave as closed pipes.
Let $AC=BC=l$
In closed pipe, the modes of vibration are given by

$l=(2n_1+1)\frac{\lambda }{4}$

${\lambda }_1=\frac{4l_1}{2n_1+1}$

Similarly, ${\lambda }_2=\frac{4l_2}{2n_2+1}$

In both gases, the frequency is same.
$\therefore f_1=f_2$

$\Rightarrow \frac{v_1}{{\lambda }_1}=\frac{v_2}{{\lambda }_2}$

$\Rightarrow \frac{v_1}{v_2}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{l_1}{l_2}\frac{2n_2+1}{2n_1+1}$

But, $l_1=l_2=0.5m$

$\Rightarrow \frac{v_1}{v_2}=\frac{2n_2+1}{2n_1+1}$

$\Rightarrow \frac{1100}{300}=\frac{2n_2+1}{2n_1+1}$

$\Rightarrow \frac{2n_1+1}{2n_2+1}=\frac{3}{11}$

For minimum frequency$n_1$ and $n_2$ should be least.

$2n_1+1=3$ and $2n_2+1=11$

$\Rightarrow n_1=1$ and $\Rightarrow n_2=5$
$\therefore f_1=(2n_1+1)\frac{v_1}{4l}=(2\times 1+1)\frac{1100}{4\times 0.5}=\frac{3\times 1100}{2}=1650Hz$