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Jalo
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Homework Statement
A lighthouse emits a noise with a frequency of 2000 Hz in the direction of the sea, with a power of 100W. Consider the speed of sound, vs, to be 341 m/s.
A ship approaches the coast with a speed of 30 km/h and produces noise with an intensity of 40dB. Consider it is possible to distinguish two sounds with equal levels of intensity. Find the maximum distance from the coast where the ship can distinguish the sound of the lighthouse.
Homework Equations
IDB = 20log10(P/P0) , with P0 = 2*10-5 N m-2
ψ(x,t) = ψ0cos(wt-kx)
ρ(x,t) = kρ0ψ0cos(wt-kx+pi/2)
P(x,t) = vsωρ0ψ0cos(wt-kx+pi/2)
I = 0.5 * vsωρ02ψ02 =
= 0.5 * Ps02 / (vsρ0) , with Ps0 equal to the amplitude of P(x,t), vsωρ0ψ0
The Attempt at a Solution
It will be impossible to hear the sound of the lighthouse when the intensity of the sound is equal to 0 decibels. Therefore:
IDB = 20log10(P/P0) = 0 ⇔ P/P0 = 1 ⇔ P = P0
I also know that the intensity is equal to the power divided by the area:
I = Power / Area
Therefore, assuming the lighthouse emits in every direction I have:
I = Power / (4∏r^2) , where r will be the distance from the lighthouse.
Equaling this equation with the expression of intensity that depends on pressure we get:
Power / (4∏r^2) = 0.5 * Ps02 / (vsρ0) ⇔
⇔ r^2 = 2*Power*(vsρ0) / [4∏*Ps02]
If I solve this equation I will not get to the correct result tho.
If anyone could point me in the right direction I'd really appreciate.
Thanks!