Sound waves propagation and the doppler effect

In summary, the conversation discusses the calculation of the maximum distance at which a ship approaching the coast can distinguish the sound of a lighthouse. The intensity of the sound is dependent on the power and distance, and it is possible to equate the intensity of the lighthouse's sound with that of the approaching ship. The equations used include the sound intensity level (IDB), sound pressure (P), and intensity (I), along with the speed of sound (vs) and density (ρ0). However, there is uncertainty about the reference level for the ship's intensity and how to relate sound pressure to power and intensity.
  • #1
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!
 
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  • #2
For what it's worth:

The units are confusing to me. I suppose I at the ship due to the lighthouse sound is 100W/4πd2 or 100W/2πd2 depending on whether the ground sound is absorbed or reflected.

The intensity at the boat is said to be +40dB, but I don't know what the intensity reference level is. In one webpage I found I_ref = 1e-12 W/m^2 so by that +40dB would be 1e-8 W/m^2.

So you can equate that level of intensity with what you get from the lighthouse at a distance d.

Maybe you can fill in some of the blanks.

I don't know how to relate sound pressure to either power or intensity but I don't see why that should be necessary here.
 

1. How do sound waves propagate?

Sound waves propagate through a medium, such as air, by creating areas of high and low pressure. This creates a chain reaction of compressions and rarefactions, which travel through the medium and reach our ears as sound.

2. What factors affect the speed of sound waves?

The speed of sound waves is affected by the density and elasticity of the medium they are traveling through. In general, sound travels faster through denser and more elastic materials.

3. What is the Doppler effect?

The Doppler effect is the change in frequency of a sound wave when the source of the wave is moving relative to the observer. This results in a perceived change in pitch of the sound.

4. How does the Doppler effect apply to sound waves?

The Doppler effect applies to sound waves when either the source or the observer is moving. If the source is moving towards the observer, the frequency and pitch of the sound wave will increase. If the source is moving away from the observer, the frequency and pitch will decrease.

5. How is the Doppler effect used in real life?

The Doppler effect has many practical applications, such as in weather forecasting to track the movement of storms, in police radar guns to measure the speed of moving vehicles, and in medical imaging to measure blood flow. It is also used in everyday technology, such as in car alarms or motion sensors.

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