Question

A working airplane engine makes a sound of frequency 12,000 Hz and the intensity level of 100 dB at a distance 100 m.

a) What is the pressure amplitude at this distance?
b) What is the displacement amplitude?
c) At what distance the sound intensity level drops to 50 dB?

Answer 1

a) The pressure amplitude at this distance is
`\Delta P_(max)=5.8 * 10^(3) N/m^(2)`

b) The displacement amplitude is
`s_(max)=1.74 * 10^(-4) m`

c) The distance at which the sound intensity level drops to 50 dB is
`r=89,200Km`

Step-by-step explanation:

a) We need to convert to dB to watts, so
`P_(dB)=10* log_(10)(P_(W))`, then
`P_(W)=10^{(P_(dB))/(10)= 10^{(100)/(10)} = 10^(10) W`.

Then we can find Intensity at 100m, so
`I=(P_(pro))/(4 \pi r^(2) )= (10^(10)W)/(4 \pi (100m)^(2) )=75.6 * 10^(3) W/m^(2)`.

And finally, the pressure amplitude at this distance will be
`\Delta P_(max)=√(\rho vI)=\sqrt{(1.29kg/m^(3))(343m/s)(75.6 * 10^(3) W/m^(2))}  =5.8 * 10^(3) N/m^(2)`

b) Using
`s_(max)=(\Delta P_(max))/(\rho \omega v)`, with
`\omega=2 \pi f=2 \pi (12000s^(-1))=\pi 24 * 10^(3) s^(-1)`, thus,

`s_(max)=(\Delta P_(max))/(\rho \omega v)= (5.8 * 10^(3) N/m^(2))/((1.29kg/m^(3)) (\pi 24 * 10^(3) s^(-1)) (343m/s) )= 1.74 * 10^(-4) m`

c) We need to find intensity, so
`10 log_(10) ((I)/(I_(0))) = 50`, then,

`I=10^(5) I_(0)=10^(5) * 10^(-12) W/m^(2)=1.00 * 10^(-7) W/m^(2)`.

Finally, the distance at which the sound intensity level drops to 50 dB will be

`r=\sqrt{(P_(pro))/(4 \pi I)} =\sqrt{(10^(10)W)/(4 \pi (1.00 * 10^(-7) W/m^(2)))}=89.2 * 10^(6)m=89,200Km`