Question

A vertical tube having unequal arm-lengths has water in it. A turning fork of frequency $550\text{Hz}$ can set up the air in the shorter arm into its fundamental mode of vibration and the same turning fork can set up the air in the longer arm into its first overtone of vibration. Find the lengths of the air column in both the arms.

[Neglect any end effect and assume that the speed of sound in air is $330{\text{ms}}^{-1}$. Take $L_1$ to be the length of shorter column and $L_2$ to be the length of longer column.]

• A. $L_1=20\text{cm}$, $L_2=40\text{cm}$
• B. $L_1=15\text{cm}$, $L_2=45\text{cm}$
• C. $L_1=10\text{cm}$, $L_2=30\text{cm}$
• D. $L_1=20\text{cm}$, $L_2=45\text{cm}$

Answer 1

The correct option is B $L_1=15\text{cm}$, $L_2=45\text{cm}$

Let the length of shorter tube be $L_1$ and that of the longer tube be $L_2$.
Since water in filled in it, tube behaves as a closed organ pipe in both arms.

Modes of vibration of air column in a closed organ pipe are given by
$f_n=\frac{(2n-1)v}{4L}\text{where}n=1,2,3,4,\mathrm{5......}$

For shorter arm:
From the data given in the question,
$550=\frac{330}{4\times L_1}\Rightarrow L_1=\frac{33}{4\times 55}=0.15\text{m}\text{or}15\text{cm}$

Similarly, for longer arm:
From the data given in the question,
$550=\frac{3\times 330}{4\times L_2}\Rightarrow L_2=\frac{3\times 330}{4\times 550}=0.45\text{m}\text{or}45\text{cm}$

Thus, option (b) is the correct answer.