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serious [3.7K]
3 years ago
14

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

aves in phase at 688 Hz\rm Hz. Take the speed of sound in air to be 344 m/s.Part BWhat is the shortest distance d d you need to walk forward to be at a point where you cannot hear the speakers? The forward direction is defined as being perpendicular to a line joining the two speakers and you start walking from the line that joins the two speakers. in meters.
Physics
1 answer:
mario62 [17]3 years ago
3 0

Answer:

The distance we need to walk in other not to hear the speakers for a speaker separation distance of 1 m, while walking in front of one of the speaker is 1.875 meters

Explanation:

The wavelength of the wave is obtained from the formula, v = f × λ

Where;

v = The velocity of the wave = 344 m/s

f = The frequency of the wave = 688 Hz

λ = The wavelength of the wave

λ = v/f = (344 m/s)/(688 Hz) = 1/2 meters

Therefore, for one not to be able to here the speakers, there must be destructive interference and R₁ - R₂ = λ/2 = (1/2)/2 = 1/4 m

Where R₁ and R₂ are the distances from the person to the two speakers respectively

When the distance between the two speakers = 1 meter, we have

R₁ = √(x² + d²), R₂ = √((1 - x)² + d²)

R₁ - R₂ = √(x² + d²) - √((1 - x)² + d²) = 1/4

When we walk from directly in front of one of the speakers, we get;

R₁ - R₂ = √(1 + d²) - √((1 - 1)² + d²) = 1/4

√(1² + d²) - d = 1/4

√(1² + d²)  = 1/4 + d

Square both sides gives

1² + d² = 1/16 + d/2 + d²

1²  = 1/16 + d/2

d/2 = 1 - 1/16 = 15/16

d = 2 × 15/16 = 15/8

d = 15/8 = 1.875 meters.

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