Question 269: Waves

Question

A pipe open at both ends of length 1.5 m is dipped in water at one end such that $2^{	ext {nd }}$ overtone of vibrating air column is resonating with a tuning fork of frequency 330 Hz . The length of the pipe immersed in water is (Speed of sound in air $=330 \mathrm{~m} / \mathrm{s}$ ) (Neglect end correction)

Accepted Answer

Let the length of the pipe immersed in water be ${ }_\chi$. For a pipe closed at one end, frequency of $2^{	ext {nd }}$ overtone is given as
$$ \mathrm{f}_3=\frac{5 \mathrm{v}}{4 \mathrm{~L}} $$
Where v is the speed of the sound wave and $\mathrm{L}_{	ext {is }}$ the length of vibrating air column
$$ \Rightarrow \quad \mathrm{f}_3=\frac{5 \mathrm{v}}{4(\mathrm{~L}-\mathrm{x})} $$
The $2^{	ext {nd }}$ overtone of vibrating air column is resonating with a tuning fork of frequency 330 Hz .
$$ \begin{array}{ll} 	herefore & 330=\frac{5 \mathrm{v}}{4(\mathrm{~L}-\mathrm{x})}=\frac{5 	imes 330}{4(1.5-\mathrm{x})} \ \Rightarrow & 1.5-\mathrm{x}=1.25 \ \Rightarrow & \mathrm{x}=0.25 \mathrm{~m} \end{array} $$

Answer

1 m

Answer

0.75 m

Answer

0.5 m

Answer

0.25 m

Waves

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