# Sound Intensity Help: Calculate Intensity & Level at 48m Mic Position

• hellokitty
In summary, we are given a concert loudspeaker emitting 36 W of sound power and a small microphone with a 1.0 cm^2 area located 48m away from the speaker. To find the sound intensity and sound intensity level at the position of the microphone, we can use the formulas I=P/4πr^2 and B=10log(I/Io). The distance of 48m corresponds to the radius r in the formula, which represents the surface area of a sphere where the sound power is spread.
hellokitty

## Homework Statement

A concert loudspeaker suspended high off the ground emits 36 W of sound power. A small microphone with a 1.0 cm^2 area is 48m from the speaker.

What is the sound intensity at the position of the microphone?

What is the sound intensity level at the position of the microphone?

## Homework Equations

I don't know where to begin. Which formulas do I use?
* I=P/4(pie)r^2 ?
*B(beta)= 10log( I/ Io)?

But it says the microphone is 1.0cmsquared area is 48 m? What does the distance x=48m fall into any of the formulas?

## The Attempt at a Solution

I Cannot solve this problem...help!

hellokitty said:
A concert loudspeaker suspended high off the ground emits 36 W of sound power. A small microphone with a 1.0 cm^2 area is 48m from the speaker.

What is the sound intensity at the position of the microphone?

What is the sound intensity level at the position of the microphone?

I don't know where to begin. Which formulas do I use?
* I=P/4(pie)r^2 ?
*B(beta)= 10log( I/ Io)?

What does the distance x=48m fall into any of the formulas?

hello hello kitty kitty!

(have a π: and a beta: ß or a different beta: β )

I don't know what the definitions of sound intensity and sound intensity level are , but the 48m is obvously the r in 4πr2.

The reason for the 4πr2 in the formula is that it's the surface area of a sphere of radius r …

the power P is spread over the whole surface area, so the power per area is P/4πr2.

As a scientist, it is important to have a thorough understanding of the concepts and equations involved in a problem before attempting to solve it. In this case, we are dealing with sound intensity and sound intensity level. Sound intensity is defined as the power of sound per unit area and is measured in watts per square meter (W/m^2). The equation for sound intensity is I = P/A, where P is the sound power and A is the area.

In this problem, we are given the sound power (P = 36 W) and the area of the microphone (A = 1.0 cm^2 = 0.0001 m^2). However, we also need to take into account the distance between the speaker and the microphone, as sound intensity decreases with distance. The formula for this is I = P/(4πr^2), where r is the distance between the speaker and the microphone in meters.

Therefore, to calculate the sound intensity at the position of the microphone, we can plug in the given values into the equation:

I = 36 W / (4π * (48 m)^2) = 0.005 W/m^2

To calculate the sound intensity level, we can use the formula B = 10log(I/Io), where Io is the reference sound intensity of 10^-12 W/m^2. This will give us the sound intensity level in decibels (dB). Again, we can plug in the previously calculated value for sound intensity (I = 0.005 W/m^2) into the equation:

B = 10log(0.005/10^-12) = 84.77 dB

Therefore, the sound intensity at the position of the microphone is 0.005 W/m^2 and the sound intensity level is 84.77 dB. It is important to note that the microphone's small area does not affect the calculation of sound intensity or sound intensity level, as these values are independent of the size of the receiving surface.

## 1. What is sound intensity and how is it calculated?

Sound intensity is a measure of the amount of energy that is transmitted through sound waves. It is typically calculated by taking the power of the sound wave and dividing it by the area in which it is being transmitted.

## 2. How do I calculate the sound intensity at a specific distance from the sound source?

To calculate the sound intensity at a specific distance, you will need to know the initial sound intensity, the distance from the source, and the inverse square law. The inverse square law states that the intensity of sound decreases by the square of the distance from the source.

## 3. What is the difference between sound intensity and sound level?

Sound intensity is a measure of the physical amount of energy in a sound wave, while sound level is a measure of the perceived loudness of a sound. Sound level takes into account the sensitivity of the human ear and is measured in decibels (dB).

## 4. How do I convert sound intensity to sound level?

To convert sound intensity to sound level, you will need to use a logarithmic formula. The most common formula is to take the logarithm of the sound intensity and multiply it by 10. This will give you the sound level in decibels (dB).

## 5. What factors can affect sound intensity?

There are several factors that can affect sound intensity, including the power of the sound source, the distance from the source, the medium through which the sound is traveling, and any obstacles or barriers that may be present. Environmental factors such as temperature and humidity can also have an impact on sound intensity.

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