Resonance Tube Experiment to Determine Speed of Sound

To determine the correct statements regarding the resonance air column experiment, we proceed with step-by-step calculations.

1. Calculation of Wavelength ($\lambda$)

In a resonance tube (an air column closed at one end), the distance between two consecutive resonance lengths corresponds to half a wavelength ($\frac{\lambda}{2}$):

$\frac{\lambda}{2} = l_2 - l_1$

Given:

• $l_1 = 50.7\text{ cm}$
• $l_2 = 83.9\text{ cm}$

$\frac{\lambda}{2} = 83.9\text{ cm} - 50.7\text{ cm} = 33.2\text{ cm}$

$\lambda = 2 \times 33.2\text{ cm} = 66.4\text{ cm} = 0.664\text{ m}$

Thus, Option C is correct.

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2. Calculation of Speed of Sound ($v$)

The speed of sound in air is related to frequency $f$ and wavelength $\lambda$ by the wave equation:

$v = f \lambda$

Given $f = 500\text{ Hz}$ and $\lambda = 0.664\text{ m}$:

$v = 500 \times 0.664 = 332\text{ m s}^{-1}$

Thus, Option A is correct.

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3. Determination of Harmonic and End Correction ($e$)

For an air column closed at one end, the resonance lengths incorporating an end correction $e$ are given by:

$l_n + e = (2n - 1) \frac{\lambda}{4}, \quad \text{where } n = 1, 2, 3, \dots$

First, we calculate $\frac{\lambda}{4}$:

$\frac{\lambda}{4} = \frac{66.4\text{ cm}}{4} = 16.6\text{ cm}$

• Testing $n = 1$ (Fundamental Harmonic): $l_1 + e = 16.6\text{ cm} \implies 50.7 + e = 16.6 \implies e = -34.1\text{ cm}$ Since $l_1 = 50.7\text{ cm} > 16.6\text{ cm}$, the resonance at $50.7\text{ cm}$ cannot be the fundamental harmonic. Thus, Option D is incorrect.

• Testing $n = 2$ (3rd Harmonic / 1st Overtone): $l_1 + e = 3 \times \left(\frac{\lambda}{4}\right) = 3 \times 16.6\text{ cm} = 49.8\text{ cm}$ $e = 49.8\text{ cm} - 50.7\text{ cm} = -0.9\text{ cm}$

Since $e = -0.9\text{ cm}$, the end correction is $-0.9\text{ cm}$ rather than $0.9\text{ cm}$. Thus, Option B is incorrect.

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Conclusion

The correct options are A and C.