Question

A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency $512Hz$ produces the first resonance when the tube is filled with water to a mark $11cm$ below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency $256Hz$ which produces the first resonance when water reaches a mark $27cm$ below the reference mark. The velocity of sound in air, obtained in the experiment, is close to

• A. $322m{s}^{-1}$
• B. $341m{s}^{-1}$
• C. $335m{s}^{-1}$
• D. $328m{s}^{-1}$

Answer 1

The correct option is D

$328m{s}^{-1}$

Step 1. Given data

Frequency of the first tuning fork, $f_1=512Hz$

Frequency of the second tuning fork, $f_2=256Hz$

The length of the first resonance for $f_1=512Hz$ is $L_1=11cm$

The length of the first resonance for $f_2=256Hz$ is $L_2=27cm$

Step 2. Finding the velocity of the sound, $v$

The first resonance frequency of length $L+e=\frac{\lambda }{4}$ [Where $e$ is the end correction and $\lambda$ is the wavelength of the sound wave.]

Using the above formula for $f_1=512Hz$, we get

$\frac{\lambda }{4}=L_1+e$

$\frac{\lambda }{4}=0.11+e$

$\frac{v}{4\times f_1}=0.11+e$ [$\lambda =\frac{v}{f}$]

$\frac{v}{4\times 512}=0.11+e$ [equation $\left(1\right)$]

Similarly for $256Hz$, we get

$\frac{v}{4\times f_2}=L_2+e$

$\frac{v}{4\times 256}=0.27+e$ [equation $\left(2\right)$]

Subtracting both equations, we get

$\frac{v}{1024}-\frac{v}{2048}=0.27+e-0.11-e$

$v=328m{s}^{-1}$

Hence, the correct option is $\left(D\right)$.