Answer
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Hint: When two longitudinal waves of same frequency travel through a medium in the opposite directions standing waves are produced.
Complete solution:
In an organ pipe column the sound behaves as a standing wave. The waves resonate harmonically.
There are two types of organ pipes;
Closed organ pipe: One end of the pipe is closed.
Open organ pipe: Both the ends are open.
In an air column filled with water, the conditions are the same as a closed organ pipe.
Hence, we can treat the situation as a closed organ pipe.
Now for standing waves in closed pipe the condition is given by;
$l = (2n + 1)\dfrac{\lambda }{4}$ (equation: 1)
Here l=length of air column, n= 0, 1, 2… $\lambda $=wavelength of the wave.
The diagram below will help you understand the situation
Now the above figure is just an example but not the actual case, as we do not know the number of harmonics.
Now, let us assume that the two successive harmonics at 50.7 and 83.9 cms be ${n_1}$ and ${n_2}$ respectively.
As the difference in length of two successive resonating air columns is half the wavelength of the wave,
$83.9 - 50.7 = \dfrac{\lambda }{2}$
Thus,$\lambda = 2 \times 33.2 = 66.4$
Hence, wavelength of sound is 66.4cm
Now as $\dfrac{\lambda }{4} = 16.6cm$, 50.7 is not the fundamental harmonic.
Also, as $\dfrac{{3\lambda }}{4} = 49.8 = e + 50.7cm$ (here, e is the end correction)
50.7 is the third harmonic and so 83.9 will be the fifth harmonic.
Now, $e = 49.8 - 50.7 = - 0.9cm$.
Hence the end correction is 0.9cm.
Now for the speed of sound $v = \lambda \upsilon $ ($\lambda $is the wavelength, $\upsilon $ is the frequency)
$v = 66.4 \times 500 = 33200cm{s^{ - 1}} = 332m{s^{ - 1}}$
Thus option A, option B and option C are correct.
Note:
(1) In the case of air columns, always consider some end correction.
(2) For organ pipe end correction may not be considered.
Complete solution:
In an organ pipe column the sound behaves as a standing wave. The waves resonate harmonically.
There are two types of organ pipes;
Closed organ pipe: One end of the pipe is closed.
Open organ pipe: Both the ends are open.
In an air column filled with water, the conditions are the same as a closed organ pipe.
Hence, we can treat the situation as a closed organ pipe.
Now for standing waves in closed pipe the condition is given by;
$l = (2n + 1)\dfrac{\lambda }{4}$ (equation: 1)
Here l=length of air column, n= 0, 1, 2… $\lambda $=wavelength of the wave.
The diagram below will help you understand the situation
Now the above figure is just an example but not the actual case, as we do not know the number of harmonics.
Now, let us assume that the two successive harmonics at 50.7 and 83.9 cms be ${n_1}$ and ${n_2}$ respectively.
As the difference in length of two successive resonating air columns is half the wavelength of the wave,
$83.9 - 50.7 = \dfrac{\lambda }{2}$
Thus,$\lambda = 2 \times 33.2 = 66.4$
Hence, wavelength of sound is 66.4cm
Now as $\dfrac{\lambda }{4} = 16.6cm$, 50.7 is not the fundamental harmonic.
Also, as $\dfrac{{3\lambda }}{4} = 49.8 = e + 50.7cm$ (here, e is the end correction)
50.7 is the third harmonic and so 83.9 will be the fifth harmonic.
Now, $e = 49.8 - 50.7 = - 0.9cm$.
Hence the end correction is 0.9cm.
Now for the speed of sound $v = \lambda \upsilon $ ($\lambda $is the wavelength, $\upsilon $ is the frequency)
$v = 66.4 \times 500 = 33200cm{s^{ - 1}} = 332m{s^{ - 1}}$
Thus option A, option B and option C are correct.
Note:
(1) In the case of air columns, always consider some end correction.
(2) For organ pipe end correction may not be considered.
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