Question

Two small speakers, 0.695 m apart, are facing in the same direction. They are driven by one 668 Hz oscillator and therefore emit identical sound waves in phase with one another at the respective points of origin. (The speed of sound waves in air is 343 m/s.)

(a) A sound engineer wishes to stand in front of one of the speakers, at the closest point (i.e., smallest x-value) where intensity is at a relative maximum. At what distance x from the nearest speaker should he position himself? (Enter your answer in m.)

(b) The sound engineer now wishes to stand at the closest point along that line where intensity is at a relative minimum. At what distance x should he position himself now? (Enter your answer in m.)

Answer 1

A) The distance x from the nearest speaker when intensity is at maximum  ; 0.4272 m

B) The distance x from the nearest speaker when intensity is at minimum ; 0.8123 m

Given data :

Distance between speakers ( d ) = 0.695 m

Frequency of oscillator = 668 Hz

Speed of sound waves in air ( v ) = 343 m/s

A) Determine the distance x from the nearest speaker when he stands where intensity is at maximum

Given that

λ = v / f

  = 343 / 668

  = 0.5135

Also

Δx = $\sqrt{d^2+x^2} - x$

Δx = λ  for maximum intensity

λ  = $\sqrt{d^2+x^2} - x$

x = ( d² - λ² ) / λ

  = ( 0.695² - 0.5135² ) / 0.5135

  = 0.4272 m

B) Determine the distance x from the nearest speaker when he stands where intensity is at minimum

Given that

λ = v / f

  = 343 / 668

  = 0.5135

Also

Δx = S₁P - S₂P

Δx = $\sqrt{d^2+x^2} - x$

since the intensity is minimum

Δx = λ/2

Therefore

λ/2  = $\sqrt{d^2+x^2} - x$

x = ( d² - λ²/4 ) / λ

  = ( 0.695² - 0.5135² / 4 ) /  0.5135

  = ( 0.695² - 0.0659 ) / 0.5135

  = 0.4171 / 0.5135 = 0.8123 m

Hence we can conclude that The distance x from the nearest speaker when intensity is at maximum  ; 0.4272 m  and The distance x from the nearest speaker when intensity is at minimum ; 0.8123 m.

Learn more about sound intensity : https://brainly.com/question/17062836