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

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.

i think its c if i'm wrong