Two loudspeakers are placed above and below each other, as in the figure below,

Question

Two loudspeakers are placed above and below each other, as in the figure below, and driven by the same source at a frequency of 4.00  102 Hz. An observer is in front of the speakers (to the right) at point O, at the same distance from each speaker. What minimum vertical distance upward should the top speaker be moved to create destructive interference at point O? (Let h = 2.78 m.)

I've tried this problem 3 different times and none of my answers were correct...I got 2.74 and 2.68 the last two times I tried...Please help...

Two loudspeakers are placed above and below each other, as in the figure below, and driven by the same source at a frequency of 4.00 102 Hz. An observer is in front of the speakers (to the right) at point O, at the same distance from each speaker. What minimum vertical distance upward should the top speaker be moved to create destructive interference at point O? (Let h = 2.78 m.)

Explanation / Answer

400Hz.
Wavelength = (v/f) = 343/400, = 0.8575 metre.
Divide by 2, = 0.42875 metre (rounded).
Path length to each speaker = sqrt.((8^2 + (2.78/2)^2)), = 8.119 metres.
(8.119 + 0.42875) = 8.5486 metres.
Sqrt.(8.5486^2 - 8^2) = 3.0130 metres.
Subtract (2.78/2), = increase in height of 1.623 metres (2 decimal places).

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