Absolute sound intensity question

[TheJourney]

Advertisement - Members don't see this ad

If the absolute sound intensity from firing the bullet is 2.0 W/m^2 at a distance of 2 m from the rifle, what will be the absolute sound intensity at a distance of 100 m from the rifle?

A. 2.0 x ^-4 W/m^2
B. 8.0 x 10^-4 W/m^2
c. 2.0 x 10^-2 W/m^2
D. 2.8 x 10^-1 W/m^2

The answer is: B (highlight to see)
I don't really understand their explanation, does anyone know the formula emlpoyed to arrive at the correct answer?

 

[el_duderino]

100 m is 50 times as far away as 2 m. Since the sound pressure is spreading over a square area (the intensity is given per unit area), the intensity at 100 m will be (intensity at 2 m) / (50^2).

2.0 W/m^2 * (2m / 100m)^2

 

[TheJourney]

I see, but how did you know to square and not simply divide by 1/50, what's the equation that shows to square?

 

[el_duderino]

Because distance is in meters and sound pressure is given in watts per meter squared. This is an inverse square law question. There's an equation for it (intensity is proportional to the inverse square of the distance, or I1/I2 = D2^2 / D1^2). The equation gives you:

2 W/m^2 / I2 = (100m)^2 / (2m)^2
I2 = 0.0008 W/m^2

But the clue is really distance vs area. As the distance increases, the sound pressure will expand over a 2-dimensional area. It'll be 1/4 the intensity at 4m than at 2m (1x1 area becomes 2x2 area), and 1/4 again (2x2 area becomes 4x4 area) at 8m than 4m.

 

[Czarcasm]

I see, but how did you know to square and not simply divide by 1/50, what's the equation that shows to square?

These types of proportionality and effect questions are very common on the MCAT. To answer this question, you'd first need to realize what intensity is. Intensity is directly proportional to power, but inversly proportional to distance squared: I=power/4(pi)r^2. In your question, focus on what's changing. We're told the distance goes from 2m to 100m - so in your head, immediately realize that it's 50 times as far (and being further, you should expect it to be less). Now taking advantage of the fact that the relationship of intensity to distance is 1/r^2 (they're inversly proportional), then you should realize that because: r^2 = 2500 (2500x as far), we should therefore expect an intensity 2500 times less than what it was originally.

If it was 2 originally then: 2 / 2500.
Calculating this, you can find the new intensity: 2 / 2.5 x10^3 ==> 2/ (5/4) x 10^3 = 0.8 x 10^-3 ==> 8x10^-4

As far as realizing the relationship to intensity was proportional to 1/r^2 (instead of just r), you had to be familiar with the relationship. But you could have reasoned it out intuitively as well. Think about a loudspeaker at a concert. The power of the loudspeaker is constant and the closer you are, the more intense it is. The further you are, the less intense. Then you should realize that the sound is radiating not in 1 dimension, but in multiple dimensions (like a sphere instead of a circle). This relationship would indication an inverse proportionality by 1/r^2.

 

[DrknoSDN]

what's the equation that shows to square?

http://en.wikipedia.org/wiki/Sound_intensity#Spatial_expansion

I2 is sound intensity at new distance.
I1 is intensity at original distance.

I2 = 2W * (2m^2 / 100m^2) = 0.0008

 

[TheJourney]

Ahhhhhhh thank you all!!! I really appreciate it!

 

[30somethingyearold]

I like how you wrote the 'highlight to see' bit - great idea!

 

[premed1001]

http://en.wikipedia.org/wiki/Sound_intensity#Spatial_expansion

I2 is sound intensity at new distance.
I1 is intensity at original distance.

I2 = 2W * (2m^2 / 100m^2) = 0.0008

do we need to know this? i dont see it on the equations list

 

[DrknoSDN]

I would say so, but the equation is quite simple. Intensity is the square of the relative distance.

Physical Science topics list has "Intensity of Sound" listed under the Sound section.
So yeah, double the distance and only have 1/4 the intensity, pretty important.
The question also comes up frequently in practice problems for decibel levels. If an sound source moves 10x farther away, how many decibels does it drop? etc.

[Edited] Hah, thanks jonnythan. Was being careless =/

 

[el_duderino]

I would say so, but the equation is quite simple. Intensity is the square of the relative distance.

Physical Science topics list has "Intensity of Sound" listed under the Sound section.
So yeah, double the distance and only have 1/2 the intensity, pretty important.
The question also comes up frequently in practice problems for decibel levels. If an sound source moves 10x farther away, how many decibels does it drop? etc.

Nitpick: double the distance, 1/4 the intensity.

 

[el_duderino]

[Edited] Hah, thanks jonnythan. Was being careless =/

SOME DOCTOR YOU'LL MAKE. WHAT IF THAT WAS A PERSON.

😉