Problem 62
Question
The sound intensity level at a rock concert is \(115 \mathrm{~dB}\), while that at a jazz fest is \(95 \mathrm{~dB}\). Determine the ratio of the sound intensity at the rock concert to that at the jazz fest.
Step-by-Step Solution
Verified Answer
The sound intensity at the rock concert is 100 times greater than that at the jazz fest.
1Step 1: Understanding the Decibel Formula
The sound intensity level in decibels (dB) is calculated using the formula \( L = 10 \, \log_{10} \left( \frac{I}{I_0} \right) \), where \( L \) is the sound level in dB, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity (typically \( 10^{-12} \, \text{W/m}^2 \)).
2Step 2: Setting Up the Equation for Both Sounds
Given, the sound intensity level at the rock concert \( L_1 = 115 \, \text{dB} \) and at the jazz fest \( L_2 = 95 \, \text{dB} \). We can express the intensity \( I_1 \) at the rock concert as \( I_1 = I_0 \times 10^{11.5} \) and \( I_2 \) at the jazz fest as \( I_2 = I_0 \times 10^{9.5} \).
3Step 3: Calculating the Intensity Ratio
The ratio of the sound intensity at the rock concert to that at the jazz fest is \( \frac{I_1}{I_2} = \frac{I_0 \times 10^{11.5}}{I_0 \times 10^{9.5}} \). Simplifying gives \( \frac{I_1}{I_2} = 10^{11.5-9.5} = 10^2 \).
4Step 4: Final Calculation and Conclusion
Calculating \( 10^2 \) gives us 100. Thus, the sound intensity at the rock concert is 100 times greater than that at the jazz fest.
Key Concepts
DecibelsIntensity RatioLogarithmic Scale
Decibels
Decibels are a unit of measurement used to express the intensity level of sound. The decibel scale is logarithmic, which means it does not increase in a linear fashion. Instead of moving by addition like a regular scale, it multiplies. This makes it particularly useful for measuring sound, which can vary over an enormous range.
The decibel scale is based on the power of ten. A change of 10 dB represents a tenfold change in intensity. For example, a sound that measures 100 dB is not twice as intense as a sound at 90 dB; it is actually ten times more intense.
The decibel scale is based on the power of ten. A change of 10 dB represents a tenfold change in intensity. For example, a sound that measures 100 dB is not twice as intense as a sound at 90 dB; it is actually ten times more intense.
- This logarithmic property helps in managing such large numbers in a more practical manner.
- It allows for a more convenient way to compare two different sound levels.
Intensity Ratio
The intensity ratio is a way of comparing two different sound intensities by forming a relative measure. It's the ratio of one sound's intensity to another's, often expressed as a power of ten due to the logarithmic nature of the scale.
In practical terms, this allows us to say how much more intense one sound is compared to another. As seen in the example of the rock concert and jazz fest, we compute the intensity ratio by using their decibel levels:
In practical terms, this allows us to say how much more intense one sound is compared to another. As seen in the example of the rock concert and jazz fest, we compute the intensity ratio by using their decibel levels:
- First, we convert each decibel reading back to its base intensity by reversing the decibel formula.
- Then, we divide one intensity by the other to get the ratio.
Logarithmic Scale
A logarithmic scale is a scale of measurement that uses the logarithm of a quantity instead of the quantity itself. This type of scale is particularly effective for data that cover a large range.
The logarithmic scale transforms exponential numbers into a manageable form. In sound:
The logarithmic scale transforms exponential numbers into a manageable form. In sound:
- The decibel formula involves a logarithmic calculation because it handles different sound intensities, which can differ by several orders of magnitude.
- By using a logarithmic scale, humans can handle and perceive changes in sound more easily, which is crucial given the way our ears perceive sound.
Other exercises in this chapter
Problem 60
A recording engineer works in a soundproofed room that is \(44.0 \mathrm{~dB}\) quieter than the outside. If the sound intensity in the room is \(1.20 \times 10
View solution Problem 61
When a person wears a hearing aid, the sound intensity level increases by \(30.0 \mathrm{~dB}\). By what factor does the sound intensity increase?
View solution Problem 63
The equation \(\beta=(10 \mathrm{~dB}) \log \left(I / I_{0}\right),\) which defines the decibel, can be written in terms of power \(P\) (in watts) rather than i
View solution Problem 65
Two identical rifles are shot at the same time, and the sound intensity level is \(80.0 \mathrm{~dB}\). What would be the sound intensity level if only one rifl
View solution