To determine which statements are true, we'll analyze the experiment step by step, using the theory of resonating air columns in a closed tube (closed at one end and open at the other).

Fundamentals of Resonance in a Closed Tube

For a tube closed at one end, the resonant lengths occur at:

$ L_n = \left( \frac{2n - 1}{4} \right) \lambda + e $

where:

$ L_n $ is the length of the air column for the $ n^\text{th} $ resonance,

$ \lambda $ is the wavelength of the sound wave,

$ e $ is the end correction,

$ n = 1, 2, 3, \dots $

Successive resonances (like those at $ L_1 $ and $ L_2 $) can be used to eliminate the end correction and find the wavelength.

Calculations

Step 1: Calculate the Wavelength ($ \lambda $)

Given:

First resonance length ($ L_1 $) = 50.7 cm

Second resonance length ($ L_2 $) = 83.9 cm

The difference between successive resonances in a closed tube is:

$ L_2 - L_1 = \frac{\lambda}{2} $

Solving for $ \lambda $:

$ \lambda = 2(L_2 - L_1) = 2(83.9\, \text{cm} - 50.7\, \text{cm}) = 2(33.2\, \text{cm}) = 66.4\, \text{cm} $

Conclusion: The wavelength of the sound wave is 66.4 cm.

Option C is true.

Step 2: Calculate the Speed of Sound ($ v $)

Using the relationship $ v = f\lambda $:

Frequency ($ f $) = 500 Hz

Wavelength ($ \lambda $) = 66.4 cm = 0.664 m

$ v = f\lambda = 500\, \text{Hz} \times 0.664\, \text{m} = 332\, \text{m/s} $

Conclusion: The speed of sound determined from the experiment is 332 m/s.

Option A is true.

Step 3: Calculate the End Correction ($ e $)

Assuming the first resonance corresponds to the second harmonic ($ n = 2 $):

$ L_1 = \left( \frac{2n - 1}{4} \right) \lambda + e $

For $ n = 2 $:

$ L_1 = \left( \frac{3}{4} \right) \lambda + e $

Solving for $ e $:

$ e = L_1 - \left( \frac{3}{4} \right) \lambda = 50.7\, \text{cm} - \left( \frac{3}{4} \times 66.4\, \text{cm} \right) $

$ e = 50.7\, \text{cm} - 49.8\, \text{cm} = 0.9\, \text{cm} $

Conclusion: The end correction is 0.9 cm.

Option B is true.

Step 4: Determine if $ L_1 $ is the Fundamental Harmonic

Assuming $ L_1 $ corresponds to the fundamental harmonic ($ n = 1 $):

$ L_1 = \left( \frac{1}{4} \right) \lambda + e $

$ e = L_1 - \left( \frac{1}{4} \right) \lambda = 50.7\, \text{cm} - 16.6\, \text{cm} = 34.1\, \text{cm} $

An end correction of 34.1 cm is unreasonably large for typical experiments, indicating that $ L_1 $ does not correspond to the fundamental harmonic.

Conclusion: The resonance at 50.7 cm does not correspond to the fundamental harmonic.

Option D is false.