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