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?
$[$Take speed of sound $=340\text{m/s}]$

• A. $45\text{cm}$
• B. $30\text{cm}$
• C. $25\text{cm}$
• D. $20\text{cm}$

Answer 1

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

We know that $v=f\lambda$
$\Rightarrow \lambda =\frac{v}{f}=\frac{340}{340}=1\text{m}$
For a closed organ pipe (organ pipe filled with water), it's fundamental resonating length,
$l_1=\frac{\lambda }{4}=\frac{1}{4}\text{m}=25\text{cm}$
Second resonating length,
$l_2=\frac{3\lambda }{4}=75\text{cm}$
Third resonating length,
$l_3=\frac{5\lambda }{4}=125\text{cm}$
Third resonance is not possible because the length of tube is only $120\text{cm}$.
$\therefore$ Minimum height of water necessary for resonance,
$H_{min}=120-75=45\text{cm}$
$$\begin{array}{c}\text{Why this question?}\\ \text{Tips: The free surface of water will form the closed}\\ \text{end of organ pipe. Resonating length is corresponding}\\ \text{to standing wave formation.}\end{array}$$