Salvatore Boone

2022-01-10

Unless indicated otherwise, assume the speed of sound in air to be $v=344\frac{m}{s}$. 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 20-mg mosquito would have to fly (in $\frac{mm}{s}$) to have this amount of kinetic energy.

Navreaiw

a) Our goal is to find the intensity of the sound wave, then to make use of the value of the intensity to determine the energy being delivered to the eardrum each second.
We can calculate the intensity of the sound wave using the relation that describes the intensity level of a sound wave, which is
$\beta =10\mathrm{log}\left(\frac{I}{{I}_{0}}\right)$ (1)
Where $\beta =20dB,{I}_{0}={10}^{-12}\frac{W}{{m}^{2}}$ and I is the intensity of the sound wave that we would like to determine.
Hence
$20dB=10\mathrm{log}\left(\frac{I}{{10}^{-12}\frac{W}{{m}^{2}}}\right)$
$2=\mathrm{log}\left(\frac{I}{{10}^{-12}\frac{W}{{m}^{2}}}\right)$
Remember that ${10}^{\mathrm{log}\left(x\right)}=x$, hence
${10}^{2}={10}^{\mathrm{log}\left(\frac{I}{{10}^{-12}\frac{W}{{m}^{2}}}\right)}$
$\frac{I}{{10}^{-12}\frac{W}{{m}^{2}}}={10}^{2}$
$I={10}^{-10}\frac{W}{{m}^{2}}$
But we know that intensity is energy per unit time per unit area (Power per unit area), which could be represented mathematically as follows
$E=IA\mathrm{△}$ (2)
Where A is the area of the tympanic membrane and it can be calculated as follows
$A=\pi {r}^{2}=\pi ×{\left(4.2×{10}^{-3}m\right)}^{2}=5.54×{10}^{-5}{m}^{2}$
Now, substitute into (2) with $5.54×{10}^{-5}{m}^{2}$ for $A,{10}^{-10}\frac{W}{{m}^{2}}$ for I and 1 s for $\mathrm{△}$ Notice that $\mathrm{△}t=1s$ since we want to calculate the energy is delivered each second
$E=\left({10}^{-10}\frac{W}{{m}^{2}}\right)×\left(5.54×{10}^{-5}{m}^{2}\right)×\left(1s\right)$
$E=5.54×{10}^{-15}J$
$l\in e\left(1,0\right)\left\{370\right\}$

Deufemiak7

b) The velocity of a mosquito with a mass of 2.0 mg and has a kinetic energy of $5.54×{10}^{-15}J$ can be calculated as follows
$KE=\frac{1}{2}m{v}^{2}$
$v=\sqrt{\frac{2×KE}{m}}$
To get v, substitute with $5.54×{10}^{-15}J$ tor KE and $2×{10}^{-6}kg$
$v=\sqrt{\frac{2×\left(5.54×{10}^{-15}J\right)}{2×{10}^{-6}kg}}$
$v=0.074×{10}^{-3}\frac{m}{s}$
$v=0.074\frac{mm}{s}$

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