Question

A tuning fork with frequency $800\, Hz$ produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonance are observed at length $9.75\, cm$, $31.25\, cm$ and $52.75\, cm$. The speed of sound in air is :

• A. $500 \,m/s^{-1}$
• B. $156 \,m/s^{-1}$
• C. $344 \,m/s^{-1}$
• D. $172 \,m/s^{-1}$

Answer 1

The Correct Option is C

To solve this problem, we need to find the speed of sound in air using the data provided about the resonance column tube. This type of problem involves understanding how resonance occurs in an air column that is closed at one end.

The fundamental principle here is that at resonance, the length of the air column corresponds to odd multiples of a quarter wavelength, i.e., the resonance occurs at:

L = \left(n + \frac{1}{2}\right)\frac{\lambda}{4}

where L is the length of the air column and \lambda is the wavelength. For a tube closed at one end, resonances occur at odd harmonics (1st, 3rd, 5th, ...).

Given the lengths of resonance as 9.75 \, \text{cm}, 31.25 \, \text{cm}, and 52.75 \, \text{cm}, we know:

1. The first resonance happens at 9.75 \, \text{cm}
2. The second resonance happens at 31.25 \, \text{cm}
3. The third resonance happens at 52.75 \, \text{cm}

Each subsequent resonance corresponds to an increase in length by \frac{\lambda}{2}. Let's calculate:

31.25 \, \text{cm} - 9.75 \, \text{cm} = 21.5 \, \text{cm}

52.75 \, \text{cm} - 31.25 \, \text{cm} = 21.5 \, \text{cm}

Thus, each increment in resonance occurs with a half wavelength. Hence, \frac{\lambda}{2} = 21.5 \, \text{cm} or \lambda = 43 \, \text{cm} = 0.43 \, \text{m}.

We know the relationship between frequency (f), wavelength (\lambda), and speed of sound (v) is given by:

v = f \times \lambda

Substituting the given values, f = 800 \, \text{Hz} and \lambda = 0.43 \, \text{m}:

v = 800 \times 0.43 = 344 \, \text{m/s}

Therefore, the speed of sound in air is 344 \, \text{m/s}, which matches the correct answer.