Question

If a resonance tube is sounded with a tuning fork of 256Hz, resonance occurs at 35cm and 105cm. The velocity of sound is about
A: 360 m/s
B: 512 m/s
C: 524 m/s
D: 400 m/s

Answer 1

Hint:Resonance tube is the best application for resonance. It is partially filled with water and a tuning fork forces it to vibration. We can deduce the answer from the relation of the resonance length of the tube and the wavelength of vibration.

Formulas used:
\[\dfrac{{{\lambda }_{{}}}}{4}={{L}_{1}}+e\] , where \[{{\lambda }_{{}}}\] is the wavelength, L1 is the length at which resonance occurs and e is the end correction.
Similarly, for the second length at which resonance occurs,
\[\dfrac{3{{\lambda }_{{}}}}{4}={{L}_{2}}+e\] , where \[{{\lambda }_{{}}}\] is the wavelength, L2 is the length at which resonance occurs and e is the end correction.

Complete step by step answer:
This is an example of a closed end organ pipe where the first harmonic or the fundamental harmonic needs to have a node at the close end since the air cannot move and an antinode at the open end.
\[\dfrac{{{\lambda }_{{}}}}{4}={{L}_{1}}+e\] …… (1)
\[\Rightarrow\dfrac{3{{\lambda }_{{}}}}{4}={{L}_{2}}+e\] …… (2)
Subtracting (1) from (2)
$\dfrac{\lambda }{2}=({{L}_{2}}-{{L}_{1}})$ ….. (3)
We know that $v=f\lambda $, where v is the velocity of sound, f is the frequency and $\lambda $ is the wavelength.
Substituting this to (3) we get
$v=2f({{L}_{2}}-{{L}_{1}})$
Substituting the given values, we get,
$\begin{align}
& v=2\times 256(105-35) \\
& \Rightarrow v=35840cm/s \\
& \therefore v\approx 360m/s \\
\end{align}$

Hence, option A is the correct answer among the given options.

Note:The first harmonic should always be considered in similar questions. The questions always insist on the fundamental mode. The first harmonic means a periodic function whose functional form is sine or cosine.