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A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally 2.00 x 10-5 W/m2 , but is turned up until the amplitude increases by 30.0%, what is the new intensity (W/m2)?

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asuwafoh

Answer:

The new intensity = 3.38 × 10⁻⁵ W/m²

Explanation:  

Intensity of sound wave:

The intensity of sound  is the rate of flow of flow of energy, per unit area, perpendicular to the direction of the sound wave.

Intensity (I) ∝ A²

where I = intensity, A = Amplitude.

∴ I₁/I₂ = A₁²/A²₂............................... equation 1

From the question, the amplitude increase by 30% of the initial

∴ A₂ = A₁ + 0.3A₁ = 1.3A₁, I₁ = 2.00×10⁻⁵ W/m²

∴ (2.00×10⁻⁵)/I₂ = A₁²/(1.3A₁)²

(2.00×10⁻⁵)/I₂ = 1/1.3²

 making I₂ the subject of the equation

I₂ = (2.00 × 10⁻⁵)×1.3² = 3.38 × 10⁻⁵ W/m₂

The new intensity = 3.38 × 10⁻⁵ W/m²

pstnonsonjoku

The new intensity of sound is 3.39  x 10-5  W/m2

Intensity of sound

We know that the intensity of sound is directly proportional to the square of the amplitude of the sound.

I α A^2

It the follows that;

I1/I2 = (A1/A2)^2

I1 =  2.00 x 10-5 W/m2

A1 = A

I2 = ?

A2 = A + 0.3A = 1.3A

Therefore;

2.00 x 10-5/I2 = (A/1.3A)^2

2.00 x 10-5/I2 = (1/1.3)^2

I2 =  2.00 x 10-5 ÷  (1/1.3)^2 =

I2 = 3.39  x 10-5  W/m2

Learn more about intensity of sound: https://brainly.com/question/3045247

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