A pair of speakers separated by distance d = 0.810 m are driven by the same osci
ID: 2254878 • Letter: A
Question
A pair of speakers separated by distance d = 0.810 m are driven by the same oscillator at a frequency of 690 Hz. An observer originally positioned at one of the speakers begins to walk along a line perpendicular to the line joining the speakers as in the figure below.
A pair of speakers separated by distance d = 0.790 m are driven by the same oscillator at a frequency of 672 Hz. An observer originally positioned at one of the speakers begins to walk along a line perpendicular to the line joining the speakers as in the figure below. How far must the observer walk before reaching a relative maximum in intensity? How far will the observer be from the speaker when the first relative minimum is detected in the intensity?Explanation / Answer
now, the idea is that you will get a maximum when two wave crests coincide; since the frequency is 690Hz and the speed of sound is 345 m/s, the wavelength is 0.5m
(use speed of wave = frequency x wavelength, solve for wavelength)
now, for the person to stand at apoint where two waves constructively interfere, the path lengths from the two speakers must differ by exactly one wavelength (or an integral multiple of wavelengths); the first such maximum occurs when the two path lengths vary by 0.5 m, this means that if the distance from the first speaker is x, the path length along the hypotenuse is x+0.5, and the pythag theorem tells us that:
x^2+0.7^2=(x+0.5)^2
solving:
x^2+0.49=x^2+x+0.25 or x=0.24
to find a minimum, where the wave destructively interfere, the path lengths must differ by a half wavelength, or by 0.25 meters, this means that if the distance from the first speaker is x, the hypotenuse is x+0.25 and Pythag tells us:
x^2+0.7^2=(x+0.25)^2
x^2+0.49=x^2+1/2 x +0.0625
1/2 x =0.49-0.0625
x=0.855 m
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