Question about Sound an intensiy

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In summary, to find the distance from a speaker where a 4000Hz sound at the same perceived loudness as a 500Hz sound at 1.0m, use the equation for intensity and the inverse square law to calculate the distance where the intensity will be 4x10^-4 W/m^2.
  • #1
cfl25
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Homework Statement



A speaker delivers 500Hz sound at an intensity of 4.0x10^-4 W/m² when measured at a distance of 1.0m from the speaker. At what distance from the speaker would you have to be to hear 4000Hz sound at the same perceived loudness?


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The Attempt at a Solution



I keep getting very short distances that do not make sense. If i could get some help on how to start this it would be great.
 
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  • #2
Intensity is proportional to the square of frequency. So find the intensity at 1m due to the source of 4000 Hz. Then using inverse square law find the distance where the intensity well be 4x10^-4 W/m^2
 
  • #3




To determine the distance at which a 4000Hz sound would be perceived at the same loudness as the 500Hz sound at a distance of 1.0m, we need to consider the inverse square law for sound intensity. The inverse square law states that the intensity of sound decreases with the square of the distance from the source. Therefore, the intensity of the 4000Hz sound at a distance of 1.0m would be significantly lower than the intensity of the 500Hz sound at the same distance.

To find the distance at which the 4000Hz sound would have the same perceived loudness as the 500Hz sound at 1.0m, we can use the formula I1/I2 = (r2/r1)^2, where I1 and I2 are the intensities of the two sounds and r1 and r2 are the distances from the source. Rearranging the equation, we get r2 = r1 x √(I1/I2).

Substituting the values given in the problem, we get r2 = 1.0m x √(4.0x10^-4 W/m² / I2), where I2 is the intensity of the 4000Hz sound.

To solve for I2, we can use the fact that the perceived loudness of sound is proportional to its intensity. Therefore, if the perceived loudness is the same for both sounds, we can set the two intensities equal to each other and solve for I2.

4.0x10^-4 W/m² = I2

Now we can substitute this value into our equation for r2:

r2 = 1.0m x √(4.0x10^-4 W/m² / 4.0x10^-4 W/m²)

r2 = 1.0m x 1

r2 = 1.0m

Therefore, the distance from the speaker at which the 4000Hz sound would have the same perceived loudness as the 500Hz sound at a distance of 1.0m is also 1.0m. This may seem counterintuitive, but it is due to the fact that the intensity of the 4000Hz sound is significantly lower than the 500Hz sound at the same distance, and thus it would need to be much closer to achieve the
 

What is sound intensity?

Sound intensity is a measure of how much energy is carried by a sound wave per unit of area. It is often described as the loudness or strength of a sound. It is measured in decibels (dB) and is directly related to the amplitude of a sound wave.

How is sound intensity measured?

Sound intensity is typically measured using a sound level meter, which detects and measures the pressure variations caused by sound waves. The results are then converted to decibels using a logarithmic scale.

What factors affect sound intensity?

The intensity of a sound can be affected by several factors, including the amplitude of the sound wave, the distance from the sound source, and the medium through which the sound travels. Other factors such as temperature, humidity, and atmospheric pressure can also play a role.

What is the difference between sound intensity and sound pressure?

Sound intensity and sound pressure are related, but not the same. Sound pressure is a measure of the force per unit area caused by sound waves, while sound intensity is a measure of the rate at which sound energy is transmitted through a unit area. In simpler terms, sound pressure is the force of the sound, while sound intensity is the strength of the sound.

Why is sound intensity important?

Sound intensity is an important concept in understanding and studying sound. It helps us quantify and compare the loudness of different sounds, and is also used to measure the potential impact of loud noises on human health and the environment. Additionally, sound intensity is used in various industries such as acoustics, engineering, and music production.

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