Physics Question -- Waves and Vibrations?

Question

Two loudspeakers are 1.00m apart. A person stands 4.00m from one speaker. How far must this person stand from the second speaker in order to detect destructive interference when the speakers emit a 1150Hz sound? Assume the speed of sound in air is 343 m/s.

Answer 1

note >> he can not stand on the line joining the speakers? think?
====================
................... A O ----- 1 m ------ O B
........................ V ................... V
................. > 4m
........................................ < x m

............................... P* person
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A & B are speakers, P person is at 4m from A, and let (x meter) from B >> so that he receives destructive interference
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interference of two waves of same freq waves reach a point of superposition> y1=a1 sin (wt+p) , y2=a2 sin wt
resultant R^2= a1^2+a2^2+2a1 a2 cos (phi)
R is max = 4a^2 (constructive with a1=a2=a), when cos phi = + 1, phase diff = o, 2pi, ....
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R is min = o (destructive with a1=a2=a), when cos phi = - 1, phase diff = pi, 3pi/2, ....
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so for destructive interference > phase differ = pi
phase diff = (2pi/ λ) path diff
pi = (2pi/ λ) path diff
path diff = λ / 2 -------- (1)
path diff = v / 2*frequency = 343/2*1150 (meter) ----- (2)
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at point P, two waves reach with path difference
path diff = x - 4 = 343/2*1150
x = 4+ [343/2*1150] = 4.149 m = 4.15 m
person should be at 4.15 m from other speaker.
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please note that person can stand anywhere on the circle of (4m) radius with centre as A (except line joining 2 speakers), but for destructive (first minima) he has to be at (4.15 m) from B >> that fixes his position wrt B.

Answer 2

For sound to travel L = vt = v/f; where v = 343 mps, L is the wavelength, and t = 1/f the period to travel one wavelength (one cycle) and f = the frequency = 1150 cycles/sec. By eyeball L ~ 1/3 meter.

Assuming the two speakers are in sync, destructive interference will be detected when standing d = n/2 L from one speaker and D = m L from the second one. m and n are integer values. There will be destruction because you'd be standing where one wave is at max compression, while the second one is at max rarification...in other words, they'd be 180 degrees out of phase, so they'd cancel out.

I should point out that d and D describe radii around their respective speakers. This is so because sound is omni directional from many if not most speakers. Thus, you can find destructive interference wherever the two circles formed by the two radii cross each other. And that happens twice for each pair of circles. So there will be destructive interference at two points for each d and D pair of values.

One final note, d and D should be close in length to each other. [d ~ D] Otherwise the energy from the closest distance to the speakers will overpower the energy from the farthest distance. In which case, you will detect only the phase of the sound wave from the closest speaker.

Answer 3

if they march at a certain pace (frequency) they could create a standing wave. think about pushing someone on a swing set, if you push them right as they stop momentarily by you, they go higher everytime.

Answer 4

Nice homework question.
Better done in the home!

Answer 5

1.1930m from the speaker
V=F~
~=ax/D
~=~ ax/D=v/f
you wanted to find the x
?