Question 49: Waves

Question

In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency $$500$$ $$Hz$$ is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length $$50.7$$ $$cm$$ and $$83.9$$ $$cm.$$ Which of the following statements is (are) true?

Accepted Answer

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} ight) \lambda + e $
where:
$ L_n $ is the length of the air column for the $ n^	ext{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 $) = 50.7Ã¢ÂÂ¯cm
Second resonance length ($ L_2 $) = 83.9Ã¢ÂÂ¯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\, 	ext{cm} - 50.7\, 	ext{cm}) = 2(33.2\, 	ext{cm}) = 66.4\, 	ext{cm} $
Conclusion:
The wavelength of the sound wave is
66.4Ã¢ÂÂ¯cm
.
Option C is true.
Step 2: Calculate the Speed of Sound ($ v $)
Using the relationship $ v = f\lambda $:
Frequency ($ f $) = 500Ã¢ÂÂ¯Hz
Wavelength ($ \lambda $) = 66.4Ã¢ÂÂ¯cm = 0.664Ã¢ÂÂ¯m
$ v = f\lambda = 500\, 	ext{Hz} 	imes 0.664\, 	ext{m} = 332\, 	ext{m/s} $
Conclusion:
The speed of sound determined from the experiment is
332Ã¢ÂÂ¯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} ight) \lambda + e $
For $ n = 2 $:
$ L_1 = \left( \frac{3}{4} ight) \lambda + e $
Solving for $ e $:
$ e = L_1 - \left( \frac{3}{4} ight) \lambda = 50.7\, 	ext{cm} - \left( \frac{3}{4} 	imes 66.4\, 	ext{cm} ight) $
$ e = 50.7\, 	ext{cm} - 49.8\, 	ext{cm} = 0.9\, 	ext{cm} $
Conclusion:
The end correction is
0.9Ã¢ÂÂ¯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} ight) \lambda + e $
$ e = L_1 - \left( \frac{1}{4} ight) \lambda = 50.7\, 	ext{cm} - 16.6\, 	ext{cm} = 34.1\, 	ext{cm} $
An end correction of
34.1Ã¢ÂÂ¯cm
is unreasonably large for typical experiments, indicating that $ L_1 $ does
not
correspond to the fundamental harmonic.
Conclusion:
The resonance at 50.7Ã¢ÂÂ¯cm does
not
correspond to the fundamental harmonic.
Option D is false.

Answer

The speed of second determined from this experiment is $$332\,m{s^{ - 1}}$$

Answer

The end correction in this experiment is $$0.9$$ $$cm$$

Answer

The wavelength of the sound wave is $$66.4$$ $$cm$$

Answer

The resonance at $$50.7$$ $$cm$$ corresponds to the fundamental harmonic

Waves

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