Question 1192174: The intensity level of sound is measured on a logarithmic scale. The intensity level, β, of sound is defined in the following equation β=10log I/Io where β is measured in decibels, and Io is the intensity of a reference level. The reference level usually taken is the "threshold of hearing", which is 1.0x10^-12 W/m^2.
a) What is the intensity level of the threshold of hearing?
b) What is the intensity level of a whisper if the intensity is 1.0x10^-10 W/m^2?
c) How much louder does a siren at 30 m away, with an intensity of 1.0x10^-2 W/m^2, sound compared to a whisper?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **Understanding Sound Intensity and Decibels**
Sound intensity is measured on a logarithmic scale using decibels (dB). This scale is based on the equation:
β = 10 log (I / I₀)
where:
* β is the sound intensity level in decibels (dB)
* I is the sound intensity in watts per square meter (W/m²)
* I₀ is the reference intensity, usually the threshold of hearing (1.0 x 10⁻¹² W/m²)
**Calculations**
**a) Threshold of Hearing**
The threshold of hearing is our reference point (I₀ = 1.0 x 10⁻¹² W/m²). Plugging this into the equation:
β = 10 log (1.0 x 10⁻¹² / 1.0 x 10⁻¹²) = 10 log (1) = 0 dB
**The intensity level of the threshold of hearing is 0 dB.**
**b) Whisper**
A whisper has an intensity of 1.0 x 10⁻¹⁰ W/m². Calculating the decibel level:
β = 10 log (1.0 x 10⁻¹⁰ / 1.0 x 10⁻¹²) = 10 log (100) = 20 dB
**The intensity level of a whisper is 20 dB.**
**c) Siren**
A siren at 30 meters has an intensity of 1.0 x 10⁻² W/m². Calculating the decibel level:
β = 10 log (1.0 x 10⁻² / 1.0 x 10⁻¹²) = 10 log (10¹⁰) = 100 dB
**The intensity level of the siren is 100 dB.**
**Comparison**
To find how much louder the siren is compared to the whisper, we subtract their decibel levels:
Difference = 100 dB - 20 dB = 80 dB
**The siren sounds 80 dB louder than the whisper.**
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