Question

At a rock concert, the sound intensity 1.0m in front of the bank of loudspeakers is 0.10 W/m^2. A fan is 30.0m from the loudspeakers. Her eardrums have a diameter of 8.4mm.How much energy is transferred to each eardrum in one second?

Answer 1

Answer: The amount of energy transferred to each eardrum in one second is $6.2 \times 10^{-9} J$.

Explanation:

Given: Intensity = 0.10 $W/m^{2}$

Now, at 30.0 m the intensity will be calculated as follows.

$Intensity = \frac{(30.0 m)^{2}}{(1 m)^{2}}\\= 900$

This means that the intensity is 900 times less assuming that the wave is spherical.

Hence, the new intensity is calculated as follows.

$\frac{0.10 W/m^{2}}{900}\\= 1.11 \times 10^{-4} W/m^{2}$

The area of ear drum is expressed as follows.

$Area = \pi \times r^{2}\\= 3.14 \times (4.2 mm)^{2}\\= 5.54 \times 10^{-5} m^{2}$

Formula used to calculate energy is as follows.

$P = I \times a$

where,

P = power or energy

I = intensity

a = area

Substitute the values into above formula as follows.

$P = I \times a\\= 1.11 \times 10^{-4} W/m^{2} \times 5.54 \times 10^{-5} m^{2}\\= 6.2 \times 10^{-9} W$

Thus, we can conclude that amount of energy transferred to each eardrum in one second is $6.2 \times 10^{-9} J$.