Q10E
Question
Question: Energy Delivered to the Ear. Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate. Typically, the diameter of this membrane is about 8.4 mm in humans. (a) How much energy is delivered to the eardrum each second when someone whispers (20 dB) a secret in your ear? (b) To comprehend how sensitive the ear is to very small amounts of energy, calculate how fast a typical 2.0-mg mosquito would have to fly (in mm/s) to have this amount of kinetic energy.
Step-by-Step Solution
Verified(a) Energy is delivered to the eardrum each second when someone whispers (20 dB) a secret in your ear is \(5.538 \times {10^{ - 15}}\;{\rm{J}}\) .
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/m2, 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 \(5.538 \times {10^{ - 15}}\;{\rm{J}}\) .