Question

A tuning fork of frequency $340\text{Hz}$ is vibrated just above the tube of $120\text{cm}$ height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance?
(speed of sound in air = $340\text{m/s}$)

• A. $45\text{cm}$
• B. $30\text{cm}$
• C. $40\text{cm}$
• D. $25\text{cm}$

Answer 1

The correct option is A $45\text{cm}$

We have

$v=f\lambda .$

or$\lambda =\frac{v}{f}=\frac{340\text{m/s}}{340\text{Hz}}=1\text{m}$

First resonating length,

$l_1=\frac{\lambda }{4}=\frac{1}{4}\text{m}=25\text{cm}$

Second resonating length,

$l_2=\frac{3\lambda }{4}=\frac{3\times 1\text{m}}{4}=75\text{cm}.$

Third resonating length,

$l_3=\frac{5\lambda }{4}=\frac{5\times 1\text{m}}{4}=125\text{cm}.$

So third resonance is not possible since the length of the tube is $120\text{cm}$.

$\therefore$ Minimum height of water necessary for resonance = $120-75=45\text{cm}$.

Why this question? For tube closed at one end the resonance occurs when the tube length is $(2n+1)\frac{\lambda }{4}$ where $n$ take values $0,1,2$....etc