91Ó°ÊÓ

The given data can be listed below as,

• The diameter of this membrane is,\(d = 8.4\;{\rm{mm}} = 8.4 \times {10^{ - 3}}\;{\rm{m}}\).
• The sound intensity is given as,\(I = 20\;{\rm{dB}}\) .
• Time is given as,\(t = 1\;{\rm{s}}\) .

The number of particles that falls on a surface in 1 second is known as the intensity of the wave with which the incident particles are associated.

Intensity is in dB, we convert it into W/m^2,and we get \(1 \times {10^{ - 10}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\) .

The delivered energy can be calculated as,

\(E = IAt\) (1)

Here A is the cross-section area.

Cross section area can be calculated as,

\(A = \pi {\left( {d/2} \right)^2}\)

\(\)

Substitute the value in the above expression, and we get,

\(\begin{array}{c}A = \pi {\left( {8.4 \times {{10}^{ - 3}}\;{\rm{m}}/2} \right)^2}\\A = 5.538 \times {10^{ - 5}}\;{{\rm{m}}^2}\end{array}\)

Now substitute the values in equation 1, and we get,

\(\begin{array}{c}E = \left( {1 \times {{10}^{ - 10}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}} \right)\left( {5.538 \times {{10}^{ - 5}}\;{{\rm{m}}^2}} \right)\left( {1\;{\rm{s}}} \right)\\ = 5.538 \times {10^{ - 15}}\;{\rm{W}} \cdot {\rm{s}}\\ = 5.538 \times {10^{ - 15}}\;{\rm{J}}\end{array}\)

Thus, energy is delivered to the eardrum each second when someone whispers (20 dB) a secret in your ear is \(5.538 \times {10^{ - 15}}\;{\rm{J}}\) .