Q70PE - College Physics (Urone)

Question

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?

Step-by-Step Solution

Answer

The new intensity is \(3.38 \times {10^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

Step 1: Identification of the given data

The given data can be listed below as:

 

The initial value of the sound intensity is \({{\rm{I}}_{\rm{1}}} = 2.00 \times {10^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

 

The final amplitude increased by 30.0% than the initial amplitude.

Step 2: Significance of the amplitude

The amplitude is described as the maximum extent of the oscillation or vibration that is measured from the equilibrium position. The amplitude of an object is directly proportional to the root of the intensity of that object.

Step 3: Determination of the new intensity

Let the initial amplitude be \({{\rm{A}}_{\rm{1}}}\) and the final amplitude is 30.0% than the initial amplitude that is \({{\rm{A}}_{\rm{1}}} \times \frac{{130}}{{100}} = 1.3{{\rm{A}}_{\rm{1}}}\).

 

The equation of the relation between the amplitude and the intensity of the microphone is expressed as:

 

\(\begin{aligned}\frac{{{{\rm{I}}_{\rm{1}}}}}{{{{\rm{I}}_{\rm{2}}}}} = \frac{{{{\rm{A}}^{\rm{2}}}_{\rm{1}}}}{{{{\rm{A}}^{\rm{2}}}_{\rm{2}}}}\\{{\rm{I}}_{\rm{2}}} = \frac{{{{\rm{I}}_{\rm{1}}}{{\rm{A}}^{\rm{2}}}_{\rm{2}}}}{{{{\rm{A}}^{\rm{2}}}_{\rm{1}}}}\end{aligned}\) 

 

Here, \({{\rm{I}}_{\rm{1}}}\) is the initial value of the sound intensity, \({{\rm{I}}_{\rm{2}}}\) is a new intensity, \({{\rm{A}}_{\rm{1}}}\) is the initial amplitude, and \({{\rm{A}}_{\rm{2}}}\) is the final amplitude.

 

Substitute \({\rm{1}}{\rm{.3}}{{\rm{A}}_{\rm{1}}}\) for\({{\rm{A}}_{\rm{1}}}\) and \(2.00 \times {10^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\) for \({{\rm{I}}_{\rm{1}}}\) in the above equation.

 

\(\begin{aligned}{{\rm{I}}_{\rm{2}}} &= \frac{{\left( {2.00 \times {{10}^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}} \right){{\left( {1.3{{\rm{A}}_{\rm{1}}}} \right)}^2}}}{{{{\rm{A}}^{\rm{2}}}_{\rm{1}}}}\\ &= \frac{{\left( {2.00 \times {{10}^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}} \right)\left( {1.69{{\rm{A}}^{\rm{2}}}_{\rm{1}}} \right)}}{{{{\rm{A}}^{\rm{2}}}_{\rm{1}}}}\\ &= \left( {2.00 \times {{10}^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}} \right)\left( {1.69} \right)\\ &= 3.38 \times {10^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{aligned}\)

 

Thus, the new intensity is \(3.38 \times {10^{ - 5}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).