Question

Sound intensity, \(\boldsymbol{\beta},\) in decibels is defined as \(\beta=10 \log \left(\frac{I}{I_{0}}\right),\) where \(I\) is the intensity of the sound measured in watts per square metre \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) and \(I_{0}\) is \(10^{-12} \mathrm{W} / \mathrm{m}^{2},\) the threshold of hearing. a) The sound intensity of a hairdryer is \(0.00001 \mathrm{W} / \mathrm{m}^{2} .\) Find its decibel level. b) A fire truck siren has a decibel level of \(118 \mathrm{dB} .\) City traffic has a decibel level of 85 dB. How many times as loud as city traffic is the fire truck siren? c) The sound of Elly's farm tractor is 63 times as intense as the sound of her car. If the decibel level of the car is \(80 \mathrm{dB},\) what is the decibel level of the farm tractor?

Short answer

a) 70 dB; b) approximately 1995.26 times; c) 98 dB.

Step-by-step solution

Identify given values for part (a)

Given: Intensity of the hairdryer, \(I = 0.00001 \, \text{W/m}^2\) and the reference intensity, \(I_0 = 10^{-12} \, \text{W/m}^2\).

Use the decibel formula for part (a)

Apply the formula for sound intensity in decibels: \(\beta = 10 \log \left(\frac{I}{I_0}\right)\).

Calculate the decibel level for part (a)

Substitute the values: \(\beta = 10 \log \left(\frac{0.00001}{10^{-12}}\right)\). Simplify the fraction: \(\beta = 10 \log \left(10^7\right)\). Since \(\log(10^7) = 7\), the decibel level is \(\beta = 10 \times 7 = 70 \, \text{dB}\).

Identify given values for part (b)

Given: Decibel level of the fire truck siren is \(118 \, \text{dB}\) and the city traffic is \(85 \, \text{dB}\).

Find the intensity ratio for part (b)

The relationship between decibel levels and intensity is \(\beta = 10 \log \left(\frac{I}{I_0}\right)\). Use the equation \(\Delta \beta = 10 \log \left(\frac{I_{1}}{I_{2}}\right)\).

Apply the decibel difference for part (b)

Find the intensity ratio: \(118 \, \text{dB} - 85 \, \text{dB} = 33 \, \text{dB}\). Therefore, \(33 = 10 \log \left(\frac{I_{fire\, truck}}{I_{city\, traffic}}\right)\). Solve for the ratio: \(\frac{I_{fire\, truck}}{I_{city\, traffic}} = 10^{3.3} \). Calculate \(10^{3.3} \), which is approximately \(1995.26\).

Identify given values for part (c)

Given: The sound of the farm tractor is 63 times as intense as the car, and the decibel level of the car is \(80 \, \text{dB}\).

Use the intensity ratio for part (c)

If \(I_{tractor} = 63 I_{car}\), use the decibel formula: \(\beta_{tractor} = 10 \log \left(\frac{I_{tractor}}{I_0}\right)\).

Calculate the decibel level for part (c)

Substitute \(I_{tractor}\): \(\beta_{tractor} = 10 \log \left(\frac{63 I_{car}}{I_0}\right)\). Since \(\beta_{car} = 80 \, \text{dB}\), we know \(\frac{I_{car}}{I_0} = 10^8\). Then: \(\beta_{tractor} = 10 \log \left(63 \times 10^8\right)\). Simplify: \(\beta_{tractor} = 10 (\log 63 + 8)\). Calculate \(\log 63\), approximately 1.8: \(\beta_{tractor} = 10 (1.8 + 8) = 10 \times 9.8 = 98 \, \text{dB}\).

Key concepts

logarithmic functions

Logarithms are a fundamental part of mathematics used to describe quantities that change multiplicatively. In our problem, we use the logarithm function to calculate sound intensity levels in decibels.
A logarithm answers the question: 'How many times must one number be multiplied by itself to get another number?' For example, \(\text{log}_{10}100 = 2\) because \({10^2 = 100}\).
Logarithmic functions are useful in many fields like science and engineering because many phenomena change exponentially. In such cases, logarithms help us handle large numbers more easily.
In the context of sound intensity, we use the base-10 logarithm, often written as \(\text{log}\). This allows us to convert large differences in sound intensity into simpler decibel values.

sound intensity

Sound intensity, in simple terms, is the amount of energy a sound wave carries per unit area. It is measured in watts per square meter (\(\text{W/m}^2\)).
Sound intensity is an objective, physical quantity, unlike loudness, which is subjective.
For example, the threshold of hearing \(I_0\), which is the quietest sound a typical human ear can hear, is \(10^{-12} \text{ W/m}^2\). More intense sounds, like a rock concert, can be around \(1 \text{ W/m}^2\).
Understanding sound intensity helps in quantifying how much sound energy is transferred to an area, and it's crucial for calculations in acoustics, audio engineering, and environmental noise monitoring.

decibel levels

Decibel levels (dB) provide a way to express the intensity of sound logarithmically, making it easier to manage the vast range of intensities our ears can detect.
The formula for decibels \(\beta\) is:
\(\beta = 10 \text{log} \frac{I}{I_0}\)
where \(I\) is the sound intensity and \(I_0\) is the reference intensity (often the threshold of hearing).
This formula tells us how many times more intense \(I\) is compared to \(I_0\).
For example, a sound with an intensity of \(\(10^{-5} \text{ W/m}^2\)\) is 70 dB. That's because \( \beta = 10 \text{ log} \frac{10^{-5}}{10^{-12}} = 70 \text{ dB} \).
Using dB makes it easier to compare sounds with significantly different intensities by converting them into a more manageable scale.

intensity ratio

The intensity ratio tells us how much more intense one sound is compared to another.
For instance, if a fire truck siren measures 118 dB and city traffic is 85 dB, we can find the intensity ratio between them.
The formula for the difference in decibels is:
\(\text{Difference in dB} = 10 \text{ log} \frac{I_{1}}{I_{2}}\).
Here, \(\text{118 dB - 85 dB = 33 dB}\).
Solving for the ratio:
\( 33 = 10 \text{ log} \frac{I_{fire truck}}{I_{city traffic}}\)
The ratio \( \frac{I_{fire truck}}{I_{city traffic}} = 10^{3.3} \), which is around 1995.26.
This means the fire truck siren is approximately 1995 times more intense than the city traffic sound.
Understanding intensity ratios helps in comparing the relative loudness of different sounds easily.