Why Is Frequency Selection Critical in Sound Interference Experiments?

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1)
In an experiment on Interference of sound, 2 loudspeakers are 0.8m apart and are driven in phase. The microphone is 1m from the loudspeakers. The frequency of the sound is set to be 500Hz. Explain why it is unsuitable to use sound of frequency, say 200Hz or 20kHz in this experiment.

Is it because the distance between the loudspeakers and the microphone is too large? Or the separation between the loudspeakers is too small?


2)
Two waves have the same frequency. Wave 1 has an amplitude of 1cm. Wave 2 has an amplitude of 2cm and leads wave 1 by 1/4 pi .

Find the amplitude of the resultant wave.

Can I simply add 1cm and 2 cm together?


3)
A transverse progressive wave of frequency 50Hz and amplitude 3cm is traveling to the right. The progressive wave is reflected and a stationary wave is formed. Now the stationary wave is illuminated witha stroboscope set at 53Hz. Find the apparent frequency of the stationary wave.

Attempt: 50 + 53 = 103Hz
I am not sure whether it is correct.

Thank you so much.

 

[denverdoc]

For the first part, assume the speakers are both facing the same direction and the mic can be anywhere on a line one meter away and parallel to a line thru the speakers. Using geometry can you figure out how the path length would vary say from the case its directly in front of the left speaker to the one in which is in the middle? Hows does path length difference relate to interference as a function of wavelength?

2) Careful here, if the two waves are ninety degrees apart, will they both be both of maximal amplitude at the same time?

3)What if the strobe was set for 50Hz? what would you see?

 

[m2003]

1) The frequency of a wave is directly related to its wavelength, and in the case of sound waves, it also affects the perceived pitch. In this experiment, the distance between the loudspeakers and the microphone is relatively short compared to the wavelength of 500Hz sound waves. This means that the sound waves will interfere constructively and destructively, creating a clear pattern of interference. However, if the frequency is changed to 200Hz or 20kHz, the wavelength will be longer or shorter, respectively. This will result in a different interference pattern, making it difficult to accurately measure and analyze the results of the experiment. Additionally, if the frequency is too low or too high, the sound may be inaudible to the human ear, making it difficult to conduct the experiment effectively. Therefore, it is best to use a frequency that is within the audible range and has a wavelength that is suitable for the distance between the loudspeakers and the microphone.

2) No, you cannot simply add the amplitudes of the two waves together. The amplitude of the resultant wave will depend on the phase difference between the two waves, in addition to their individual amplitudes. In this case, since the waves have the same frequency, the resultant wave will have an amplitude of 3√2 cm. This can be calculated using the formula A = √(A1^2 + A2^2 + 2A1A2cosΦ), where A1 and A2 are the amplitudes of the two waves and Φ is the phase difference between them.

3) The apparent frequency of the stationary wave will not be equal to the sum of the original frequency (50Hz) and the strobe frequency (53Hz). This is because the stationary wave is a result of the interference between the original wave and its reflection, and the two waves have a phase difference of π. The apparent frequency can be calculated using the formula f' = (f1 + f2)/2, where f1 and f2 are the original frequencies and f' is the apparent frequency. In this case, f' = (50Hz + 53Hz)/2 = 51.5Hz. Therefore, the apparent frequency of the stationary wave will be 51.5Hz.