Question

In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form?

Answer 1

Given:
Length of air column at first resonance L₁ = 20 cm = 0.2 m
Length of air column at second resonance L₂ = 62 cm = 0.62 m
Frequency of tuning fork f = 400 Hz

(a) We know that:
$$\begin{gathered} \lambda =2\left(L_2-L_1\right) \\ \Rightarrow \lambda =2\left(62-20\right)=84\mathrm{cm}=0.84\mathrm{m} \end{gathered}$$

v = λf,
where v is the speed of the sound in air.
So,
$v=0.84\times 400=336\mathrm{m}/\mathrm{s}$
Therefore, the speed of the sound in air is 336 m/s.

(b) Distance of open node is d:

$$\begin{gathered} L_1+d=\frac{\lambda }{4} \\ \Rightarrow d=\frac{\lambda }{4}-L_1=21-20=1\mathrm{cm} \end{gathered}$$
Therefore, the required distance is 1 cm.