Two loud speakers are 2.63 m apart. I am standing 2.5 m from one of them and 3.5

Question

Two loud speakers are 2.63 m apart. I am standing

2.5 m from one of them and 3.58 m from the other. The two speakersare

driven by a single oscillator. If the frequency of the oscillatoris swept from 100 Hz to

1000 Hz, ?nd the lowest frequency (Hz) at which I will hear anenhancement of the sound

intensity due to constructive interference of the waves from thetwo speakers. Use

343 m/s for the speed of sound. (DO NOT use the double-slitequation, d sin = m.

This equation is only valid for observations far from the twoslits, compared to the

distance between the two slits.)

Explanation / Answer

Distance of you from one of the speaker y = 2.5 m Distance of you from second speaker y ' = 3.58 m Path difference of the two sound waves   D = y ' -y                                                                  = 1.08 m Speed of the sound v = 343 m / s wavelength of the wave = v / f here f = frequency of the wave = its value inbetween 100 Hz to 1000 Hz if the path difference is equal to wavelength thenintensity of the sound is maximum That means v / f = 1.08 m                        f = v / 1.08                          = 343 / 1.08                          = 317.59 Hz i.e., required lowest frequency f = 317.59 Hz

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