Sound wave intensity of elephants

[sun]

Homework Statement

There is evidence that elephants communicate via infrasound, generating rumbling vocalizations as low as 14hz that can travel up to 10km. The intensity level of these sounds can reach 103dB, measured a distance of 5.0m from the source. Deteremine the intensity level of the infrasound 10km from the source, assuming the sound energy radiates uniformly in all directions.

Homework Equations

[tex]B[/tex]=10log(I/10^-12)

The Attempt at a Solution

i found the intensity at an intensity level of 103dB to be.0199W/m^2. I'm just not sure where to go from here. Suggestions would be appreciated.

thanks

 

[rl.bhat]

Intensityi is inversly proportional to square of the distance.
Using I1/R1^2 = I2/R2^2, find I2. Then convert it into dB.

 

[ogb p]

I would approach this problem by first considering the properties of sound waves and how they propagate through space. Sound waves are a type of mechanical wave, meaning they require a medium to travel through. In this case, the medium is air. Sound waves also exhibit the property of inverse square law, which states that the intensity of a sound wave decreases as the distance from the source increases.

To determine the intensity level of the infrasound 10km from the source, we can use the equation B=10log(I/10^-12) as provided in the homework. Here, B represents the intensity level in decibels, I represents the intensity in watts per meter squared, and the value 10^-12 is a reference value used to compare sound intensities.

Since we are given the intensity level at a distance of 5.0m from the source, we can use this as a reference point to calculate the intensity at 10km. Using the inverse square law, we know that the intensity at 10km will be 1/40th of the intensity at 5.0m. Therefore, we can multiply the intensity at 5.0m by 1/40 to get the intensity at 10km.

Using the value of 0.0199W/m^2 for the intensity at 5.0m, we can calculate the intensity at 10km as 0.0199W/m^2 * 1/40 = 0.0004975W/m^2.

Now, to determine the intensity level at 10km, we can plug this value into the equation B=10log(I/10^-12). This gives us B=10log(0.0004975/10^-12) = 93.96dB.

Therefore, the intensity level of the infrasound at a distance of 10km from the source is approximately 93.96dB. This is significantly lower than the intensity level at 5.0m, demonstrating the effect of the inverse square law on sound propagation.