Two speakers are driven by a common oscillator at 850 Hz and face each other at

Question

Two speakers are driven by a common oscillator at 850 Hz and face each other at a distance of 1.24 m. Locate the points along a line joining the two speakers where relative minima of the amplitude of the pressure would be expected. (Use v = 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker. Enter 'none' in all unused answer boxes.)

Minima Distance from
speaker (m) 1st
Your response differs from the correct answer by more than 100%. 2nd
Your response differs from the correct answer by more than 100%. 3rd 4th 5th 6th 7th

Explanation / Answer

= 343/850 = 0.404 m

Sound amplitude due o one speaker
p1 = po sin(kx - t) and due to the opposite speaker
p2 = po sin [k(1.24 - x) - t]
The resultant sound pressure amplitude
p
= p1 + p2
= 2po sin (0.62k - t) cos (kx - 0.62k)
For p to be zero,
(x - 0.62)k = (2n - 1)/2
=> (x - 0.62) 2/ = (2n - 1)/2
=> x
= 0.62 + (2n - 1) /4
= 0.62 + (2n - 1) (0.404/4)
= 0.62 + (2n -1)(0.101)
Use values of n that gives x between 0 and 1.25.
n = 1 => x = 0.721 m
n = 2 => x = 0.923 m
n = 0 => x = 0.519 m
n = -1 => x = 0.317 m
n = -2 => x = 0.115 m

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