Intensity of periodic sound waves

In summary, at a distance of 400m from a fireworks explosion, the acoustic pressure reaches a maximum of 10.0 N/m2. The sound level (dB) at 4.00km from the explosion can be calculated using the equation \beta = 10 log \left \frac{P}{P_0} \right +br, where P is the acoustic pressure, P_0 is the reference pressure, and b is the energy loss rate of -7 dB/km. The ground absorption means that there is no reverberation or reflection, so only the acoustic pressure needs to be considered. Using the acoustic pressure of 10 at 0.4km, the acoustic pressure at 4.0km
  • #1
kreil
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The problem is this:

A firework charge is detonated many meters above the ground. At a distance of 400m from the explosion, the acoustic pressure reaches a maximum of 10.0 N/m2. Assume that the speed of sound is constant at 343 m/s throughout the atmosphere over the region considered, that the ground absorbs all the sound falling on it, and that the air absorbs sound energy as described by the rate 7.0 dB/km.

What is the sound level (dB) at 4.00km from the explosion?

I know that

[itex]\beta = 10 log \left \frac{I}{I_0} \right [/itex]

and that

[itex]I=\frac{P}{A}=\frac{1}{2} p v w^2 {s^2_{max}}[/itex]

where p is the density of air, v is the speed of sound, w is the angular frequency and smax is the amplitude of the position function s(x,t)=smaxcos(kx-wt).

but I am having trouble correctly solving for I, and so I can't get the book answer of B=65.6 dB. Any help is appreciated.

Note: the equation I obtained for this problem taking into account the damping of the sound in air is:

[itex]\beta = 10 log \frac{I}{I_0}+br[/itex]

where b=-7 dB/km and r=4.0km is the distance from the explosion.
 
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  • #2
Basically, I'm not sure how to use the acoustic pressure or ground damping in the problem.
 
  • #3
[itex]\beta = 20 log \left \frac{P}{P_0} \right [/itex]

and the ground absorption means no reverberation or reflection, so one only need to be concerned with acoustic pressure.

Is there an example of how to use the energy loss rate (db/km)?

One is given two distances, 0.4 km and 4.0 km, and the acoustic pressure at 0.4 km.
 
  • #4
What reference pressure should I use in that equation?

If the acoustic pressure at 0.4km is 10, then at 4.0km it will be 1.

It doesn't say how to use the energy loss rate, so I guessed. Now that I look at it, it doesn't look right. Any ideas?
 

FAQ: Intensity of periodic sound waves

1. What is the relationship between intensity and amplitude of a periodic sound wave?

The intensity of a periodic sound wave is directly proportional to the square of its amplitude. This means that as the amplitude of a sound wave increases, its intensity also increases.

2. How is intensity of a periodic sound wave measured?

The intensity of a periodic sound wave is measured in decibels (dB) using a logarithmic scale. This scale allows for a wider range of values to be represented, as sound waves can vary greatly in intensity.

3. What factors affect the intensity of a periodic sound wave?

The intensity of a periodic sound wave can be affected by the amplitude of the wave, the distance from the source of the sound, and the medium through which the sound is traveling. Other factors such as temperature and humidity can also have an impact on sound wave intensity.

4. Can the intensity of a periodic sound wave be changed?

Yes, the intensity of a periodic sound wave can be changed by altering the amplitude of the wave. Other methods such as using a sound amplifier or changing the distance from the sound source can also affect the intensity of a sound wave.

5. How does the intensity of a periodic sound wave affect human perception?

The intensity of a periodic sound wave can have a significant impact on human perception. Higher intensities can be perceived as louder and may cause discomfort or even damage to the ears. On the other hand, lower intensities may not be heard at all. The perceived pitch of a sound can also change with different intensities.

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