Calculate the Sound intensity level

In summary, the problem asks to calculate the sound intensity levels in dB at distances of 2.0 m and 50.0 m from a jackhammer with a sound power of 10 W. This can be done by using the formula β = (10dB) log (I/Io) where β is the sound intensity level, I is the intensity, and Io is a constant of 1.0x10^-12 W/m^2. To calculate the intensity at these distances, the surface area that the sound is going through must be determined. The appropriate surface area can be either a sphere or a hemisphere, depending on the assumptions of the problem. Once the intensity is calculated, the corresponding levels in dB can be found by
  • #1
ninaw21
15
0

Homework Statement


A jackhammer with a sound power of 10 W is operating in the countryside. Calculate the
intensity levels at distances of 2.0 m and 50.0 m from the jackhammer. Calculate the
corresponding levels in dB.


Homework Equations


β = (10dB) log (I/Io)
where, β = sound intensity level, I = intensity, Io= 1.0x10^-12 W/m^2


The Attempt at a Solution


I honestly have no idea how to do this question! Please help
 
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  • #2
Welcome to PF.

If you inspect your formula, you will notice that you need to find I. The definition of I is power per unit area: I = P / A.

Assume the jackjammer is a point source. Then A will be the surface area of a sphere. You are given the power.

Try this and show your work !
 
  • #3
ninaw21 said:

Homework Statement


A jackhammer with a sound power of 10 W is operating in the countryside. Calculate the
intensity levels at distances of 2.0 m and 50.0 m from the jackhammer. Calculate the
corresponding levels in dB.


Homework Equations


β = (10dB) log (I/Io)
where, β = sound intensity level, I = intensity, Io= 1.0x10^-12 W/m^2


The Attempt at a Solution


I honestly have no idea how to do this question! Please help

Welcome to the PF.

It looks like the units of sound intensity are W/m^2:

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

So you would want to calculate the surface area that the 10W is going through at those distances away from the source, and use that surface area and the 10W number to figure out the sound intensity at those distances.

One question would be should you use a sphere's area or hemisphere's area for your surface that the sound is going through. I'm not sure which this problem assumes. I know which one I'd use based on the wording, but I don't know if it would be the same as the problem wants...


EDIT -- Edged out by edgepflow! :smile:
 
  • #4
Thanks guys!
 
  • #5
!

Hi there,

As a scientist, it's important to have a solid understanding of sound intensity levels and how to calculate them. Let's break down this problem step by step.

First, we need to understand that sound intensity is a measure of the amount of sound energy passing through a unit area per unit time. It is measured in watts per square meter (W/m^2).

In this problem, we are given the sound power of the jackhammer, which is 10 W. This means that 10 watts of sound energy is being produced by the jackhammer every second.

To calculate the sound intensity at a distance of 2.0 m, we can use the inverse square law, which states that the intensity is inversely proportional to the square of the distance. So, at a distance of 2.0 m, the sound intensity would be 10 W / (2.0 m)^2 = 2.5 W/m^2.

To calculate the sound intensity at a distance of 50.0 m, we would use the same formula, but with a distance of 50.0 m. This would give us an intensity of 10 W / (50.0 m)^2 = 0.04 W/m^2.

Now, to calculate the sound intensity level (β), we can use the formula given in the problem: β = (10 dB) log (I/Io). Plugging in the values for I and Io from our calculations above, we get:

- At 2.0 m: β = (10 dB) log (2.5 W/m^2 / 1.0x10^-12 W/m^2) = 123 dB
- At 50.0 m: β = (10 dB) log (0.04 W/m^2 / 1.0x10^-12 W/m^2) = 107 dB

These are the sound intensity levels at the given distances. To put them in context, a normal conversation is around 60 dB, so the jackhammer would be significantly louder at both distances.

I hope this helps you understand how to calculate sound intensity levels. Don't hesitate to ask for clarification if you need it. Good luck with your studies!
 

Related to Calculate the Sound intensity level

What is sound intensity level?

Sound intensity level is a measure of the sound energy per unit area. It is expressed in decibels (dB) and is used to quantify the loudness or intensity of a sound wave.

How is sound intensity level calculated?

The sound intensity level (SIL) is calculated using the formula SIL = 10 log(I/I0), where I is the sound intensity and I0 is the reference intensity of 10^-12 watts per square meter.

What is the reference intensity used for sound intensity level?

The reference intensity of 10^-12 watts per square meter is considered to be the lowest intensity that the human ear can perceive. It is based on the threshold of hearing for a young, healthy human with average hearing.

How does sound intensity level relate to loudness?

Sound intensity level is directly related to loudness. As the intensity level increases, the perceived loudness also increases. A 10 dB increase in sound intensity level is perceived as a doubling of loudness.

What are typical sound intensity levels for common sounds?

The sound intensity levels of common sounds can vary greatly. Here are some examples: whispering (20 dB), conversation (60 dB), vacuum cleaner (70 dB), rock concert (110 dB), and a jet engine (140 dB). It is important to note that prolonged exposure to sounds above 85 dB can lead to hearing damage.

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