Question

A fan at a rock concert is 42.2m from the stage, and at this point the sound intensity level is 114.9dB . Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate . Typically, the diameter of this membrane is about 8.4mm in humans

1.How much energy is transferred to her eardrums each second ?

2.How fast would a 2mg mosquito have to fly to have this much kinetic energy?

3. How fast a typical 2.0 mg mosquito would have to fly (in mm/s) to have an amount of energy delivered to the eardrum each second when someone whispers (20 dB) a secret in your ear?

4. Compare the mosquito's speeds found in parts (B) and (C).

Answer 1

1) Sound intensity level

$\beta =10 dB \cdot log\left ( \frac{I}{I_o} \right )$

$119.4dB =10 dB \cdot log\left ( \frac{I}{1\times 10^{-12}} \right )$

$I= 0.871 W/m^2$

The energy transfered per second is

$P = IA=(0.871)\pi r^2 = (0.871)\pi (4.2\times 10^{-3})^2 = 4.8267\times 10^{-5}J$

2) the kinetic energy of the mosquito is

$4.8267\times 10^{-5}= \frac{1}{2}(2\times 10^{-6}kg)V^2$

$V = 6.95 m/s$

3)

$20dB =10 dB \cdot log\left ( \frac{I}{1\times 10^{-12}} \right )$

$I= 1\times 10^{-10} W/m^2$

The energy transfered per second is

$P = IA = (1\times 10^{-10})\pi r^2 = (1\times 10^{-10})\pi (4.2\times 10^{-3})^2 = 5.54 \times 10^{-15}J$

the kinetic energy of the mosquito is

$5.54\times 10^{-15}= \frac{1}{2}(2\times 10^{-6}kg)V^2$

$V = 0.0744 mm/s$

d)

Comparing the speed

6.95 m/s / 0.0744 mm/s = 93360 times.

The part b speed is 93360 times higher than part c speed.