Question

In a resonance tube experiment with a tuning fork of frequency 480Hz the consecutive resonance lengths are 30cm and 70cm then determine the speed of sound is
A) \[200m/s\]
B) \[256m/s\]
C) \[384m/s\]
D) \[240m/s\]

Answer 1

Hint: Resonance is a phenomenon in which a system or an object is made to oscillate about it’s mean position when an external force is applied to it. Here in order to determine, speed of sound in resonating lengths, a resonance tube experiment is performed. In this experiment, an open ended cylindrical tube is used to find the velocity of the sound in air.

Complete step by step solution:
Step I:
The velocity by which a sound wave travels in any medium can be known. For that, the frequency and wavelength of the waves should be known first. This is because the velocity is written as
\[{V_{sound}} = f\lambda \]---(i)
Where \[V\] is the velocity of the sound wave.
\[f\] is the frequency
And \[\lambda \] is the wavelength of the sound wave
Step II:
In the resonance tube experiment, the resonance of a sound wave occurs, if the length of the tube taken is a multiple of half the wavelength of sound. If \[\lambda \]is the wavelength considered then the resonating length is given by \[\dfrac{{n\lambda }}{2}\]. Let the first resonating length is 30 cm.
As discussed above,
\[\dfrac{{n\lambda }}{2} = 30cm\]---(i)
Since resonating lengths are consecutive. So the next resonating length will be \[\dfrac{{(n + 1)\lambda }}{2}\].
\[\dfrac{{(n + 1)\lambda }}{2} = 70\]---(ii)
Step II:
Solving the above equation and evaluating the value of \[\dfrac{\lambda }{2}\].
\[\dfrac{{n\lambda }}{2} + \dfrac{\lambda }{2} = 70\]
Substitute value from equation (i),
\[30 + \dfrac{\lambda }{2} = 70\]
\[\dfrac{\lambda }{2} = 40cm\]
\[\lambda = 80cm\]
\[\lambda = 0.8m\]
Step III:
Given \[f = 480Hz\]
Substituting values in equation (i),
\[{V_{sound}} = f\lambda \]
\[{V_{sound}} = 480 \times 0.8\]
\[{V_{sound}} = 384m/s\]

Hence, Option (C) is the correct answer.

Note:
In order to find the wavelength of the resultant stationary waves produced in the resonance tube experiment, the resonating length is calculated. When the length of the waves decreases, the period also decreases. This will result in a shorter frequency of the pipe used. The frequency varies directly with the length of the resonating tube.