Question

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeakers. Her eardrums have a diameter of 8.4 mm. How much sound energy is transferred to each eardrum in 1.0 second?

Answer 1

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

`P = AI`

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

`A = (\pi d^2)/(4)`

Now the intensity is inversely proportional to the square of the distance from the source, then

`I \propto (1)/(r^2)`

The expression for the intensity at different distance is

`(I_1)/(I_2)= (r^2_2)/(r_1^2)`

Here,

`I_1` = Intensity at distance 1

`I_2` = Intensity at distance 2

`r_1` = Distance 1 from light source

`r_2` = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

`I_2 = I_1 ((r_1^2)/(r_2^2))`

If we replace with our values at this equation we have,

`I_2 = (0.10W/m^2)((1.0m^2)/(30.0m^2))`

`I_2 = 1.11*10^(-4) W/m^2`

Now using the equation to find the area we have that

`A = (\pi (8.4*10^(-3)m)^2)/(4)`

`A = 5.5*10^(-5)m^2`

Finally with the intensity and the area we can find the sound power, which is

`P = AI`

`P = (5.5*10^(-5)m^2)(1.11*10^(-4)W/m^2)`

`P = 6.1*10^(-9)J/s`

Power is defined as the quantity of Energy per second, then

`E = 6.1*10^(-9)J`