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rodikova [14]
3 years ago
8

When a sound wave travels directly toward a hard wall, the incoming and reflected waves can combine to produce a standing wave.

There is an antinode right at the wall, just as at the end of a closed tube, so the sound near the wall is loud. You are standing beside a brick wall listening to a 85 Hz tone from a distant loudspeaker.How far from the wall must you move to find the first quiet spot?
Physics
1 answer:
murzikaleks [220]3 years ago
3 0

Answer:

Explanation:

When standing sound wave is produced in a medium , distance between anti-node and next node is equal to a quarter of a wave length. Node is point of silence where sound intensity is zero or minima.

wave length = velocity / frequency

= 340 / 85

= 4 m

We must  move by a distance of 4/4 = 1 m away from wall

to find the first quiet spot.

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Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi
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Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

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ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

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n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

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v = fλ

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Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

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