Question

In a resonance tube, using a tuning fork of frequency 325 Hz, two successive resonance lengths are observed as 25.4 cm and 77.4 cm respectively. The velocity of sound in air

• A. 338 ms^-1
• B. 328 ms^-1
• C. 330 ms^-1
• D. 320 ms^-1

Answer 1

The correct option is A

338 ms^-1

Step 1: Given terms

Frequency, $\mathrm{f}=325\mathrm{Hz}$

First resonance length, ${\mathrm{l}}_1=25.4\mathrm{cm}$

Second resonance length, ${\mathrm{I}}_2=77.4\mathrm{cm}$

Step 2: Formula used

The velocity of sound in the air for the resonance tube is given as,

General Form: wave motion $v=f\lambda$.

Wave motion is used to calculate the wavelength between two successive nodes.

Here $\lambda$ is the wavelength (difference of I₁, I₂)

$\mathrm{v}=2\times \mathrm{f}({\mathrm{l}}_1-{\mathrm{l}}_2)$

Here v is velocity, I₁,I₂ are two lengths and f is frequency.

Step 3: Calculating velocity

On substituting the values in the formula, we get:

$$\begin{gathered} v=2\times 325\times \left(77.4-25.4\right)\times {10}^{-2} \\ =338m{s}^{-1} \end{gathered}$$

Hence, the velocity of air is $338{\mathrm{ms}}^{-1}$.