Question

Resonance is produced between a tuning fork and a resonance column tube with its upper end open and lower end closed by a water surface. If the frequency of the tuning fork is $800\text{ Hz}$, and the first two resonances are observed at lengths $75\text{ cm}$ and $325\text{ cm}$, find the length at which the third resonance occurs and the speed of sound.

• A. $43.10\text{ cm}$ and $340\text{ m s}^{-1}$
• B. $31.25\text{ cm}$ and $330\text{ m s}^{-1}$
• C. $62.60\text{ cm}$ and $335\text{ m s}^{-1}$
• D. $52.75\text{ cm}$ and $344\text{ m s}^{-1}$

Hint:

For a closed pipe system, the gap between consecutive positions is constant. Since the gap between the first two values is $21.5\text{ cm}$, simply add $21.5$ to the second length to get the third position instantly: $31.25 + 21.5 = 52.75\text{ cm}$. This leaves only Option (D) as a viable choice!

Answer 1

The Correct Option is D

Concept: In a resonance column apparatus (acting as a closed organ pipe), consecutive resonance lengths $l_1, l_2, l_3$ are separated by half a wavelength ($\frac{\lambda}{2}$) to cancel out end correction effects: \[ l_2 - l_1 = \frac{\lambda}{2} \quad \text{and} \quad l_3 - l_2 = \frac{\lambda}{2} \] The speed of sound ($v$) is calculated using the wave equation $v = f\lambda$.

Step 1: Calculate the wavelength ($\lambda$) and find the third resonance length ($l_3$).
Given $l_1 = 9.75\text{ cm}$ and $l_2 = 31.25\text{ cm}$: \[ \frac{\lambda}{2} = 31.25 - 9.75 = 21.5\text{ cm} \implies \lambda = 43.0\text{ cm} = 0.43\text{ m} \] Since consecutive lengths increase by equal steps of $\frac{\lambda}{2} = 21.5\text{ cm}$: \[ l_3 = l_2 + 21.5 = 31.25 + 21.5 = 52.75\text{ cm} \]

Step 2: Calculate the speed of sound ($v$).
Using the given frequency $f = 800\text{ Hz}$ and $\lambda = 0.43\text{ m}$: \[ v = f\lambda = 800 \times 0.43 = 344\text{ m s}^{-1} \]