Intensity of sound, power output

In summary, the conversation discusses the calculation of intensity (I) and power (P) of sound radiation, with a specific focus on the proportion between the two. The questioner initially receives a value of 49mW for the power, but believes it should be 97.xxmW due to a mistake in using the formula for a hemisphere instead of considering the non-spherical distribution of energy. The responder clarifies the correct formula to be used and prompts the questioner to consider the impact of the non-spherical distribution on the calculation. Finally, the questioner understands and expresses gratitude.
  • #1
jangchen
12
3
Homework Statement
A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the back. You measure an 85 dB sound intensity level when standing 3.5 m in front of and 2.5 m below the speaker. What is the speaker’s power output?
Relevant Equations
dB = 10 log(I / Io) , I = P / 4pi×r^2
I is proportionate to E.

r = 4.3xxm

So I put 0.75I

The answer is 49mW..

But I got 97.xxmW

What is wrong?
 
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  • #2
I get 49 mW. If you show the details of your calculation we can maybe spot a mistake. If I had to guess, I would say you used 4πr2 for the area of a hemisphere.
 
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  • #3
Your equation for I assumes the energy distribution is spherically symmetrical. But the question tells you that is not true. If 75% of the energy is emitted forwards (as opposed to 50% for spherical symmetry), how would that affect the I value you measure?
 
  • #4
TSny said:
If I had to guess, I would say you used 4πr2 for the area of a hemisphere.
Oh, I used 4πr2.. Do I have to use 2πr2 ?
 
  • #5
If 75% rather than 50% of the energy goes forwards, then
Iforward = 1.5P/4πr2
I'm not sure what you meant by "So I put 0.75I", but if it means you used 0.75 instead of 1.5 in the above equation, then that would explain your error.
 
  • #6
There's more than one way to think about it. Can you answer each of the following?
(1) What is the intensity I (in W/m2) at the point given?
(2) What is the total power P (in W) of sound radiated in the forward direction (hemisphere)?
(3) How is the answer to (2) related to the total power radiated by the speaker?
 
  • #7
Now I understand surely. Thank you!
 

1. What is the difference between intensity of sound and power output?

The intensity of sound refers to the amount of energy that is transmitted through a certain area over a period of time. It is measured in watts per square meter. Power output, on the other hand, is the rate at which energy is produced or transferred. It is measured in watts.

2. How is intensity of sound measured?

Intensity of sound is typically measured using a decibel (dB) scale. This scale compares the sound intensity to the threshold of human hearing, which is 0 dB. A sound with an intensity of 10 dB is 10 times more intense than the threshold of hearing.

3. Does the distance from the source affect the intensity of sound?

Yes, the intensity of sound decreases as the distance from the source increases. This is because the sound energy is spread out over a larger area, resulting in a lower intensity. The inverse square law states that the intensity is inversely proportional to the square of the distance from the source.

4. How does power output affect the intensity of sound?

Power output and intensity of sound are directly related. As the power output increases, the intensity of sound also increases. This is because more energy is being produced and transferred, resulting in a higher intensity of sound.

5. How does the medium affect the intensity of sound?

The medium through which sound travels can affect its intensity. In general, sound travels with greater intensity in denser mediums, such as water or solids, compared to less dense mediums, like air. This is because the particles in denser mediums are closer together, allowing for more efficient transfer of energy.

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