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Chapter 16: Problem 62

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.

Short Answer

Expert verified
The sound intensity at the rock concert is 100 times greater than that at the jazz fest.

Step by step solution

01

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 \( 10^{-12} \, \text{W/m}^2 \)).
02

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 \( I_2 = I_0 \times 10^{9.5} \).
03

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 \( \frac{I_1}{I_2} = 10^{11.5-9.5} = 10^2 \).
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
  • 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.
Understanding decibels is essential for discussing sound intensity, as it allows us to easily express the very large or very small ratios involved in sound measurements.
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:
  • 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.
This process helps us quantify and understand differences in sound intensities in a straightforward manner. So, when we say that the rock concert is 100 times more intense than the jazz fest, we're relying on this 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 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.
Without a logarithmic scale, measuring something like sound intensity鈥攚hich can range from a pin drop to a jet engine鈥攚ould be cumbersome and imprecise. This scale turns potentially unwieldy numbers into figures that are easier to understand and use.

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Most popular questions from this chapter

The displacement (in meters) of a wave is given according to \(y=0.26 \sin (\pi t-3.7 \pi x)\) where \(t\) is in seconds and \(x\) is in meters, (a) Is the wave traveling in the \(+x\) or \(-x\) direction? (b) What is the displacement \(y\) when \(t=38 \mathrm{~s}\) and \(x=13 \mathrm{~m} ?\)

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 intensity \(I\) (in watts/meter \(^{2}\) ). The form \(\beta=(10 \mathrm{~dB}) \log \left(P / P_{0}\right)\) can be used to compare two power levels in terms of decibels. Suppose that stereo amplifier A is rated at \(P=250\) watts per channel, and amplifier B has a rating of \(P_{0}=45\) watts per channel, (a) Expressed in decibels, how much more powerful is A compared to \(\mathrm{B} ?\) (b) Will A sound more than twice as loud as \(\mathrm{B}\) ? Justify your answer.

A portable radio is sitting at the edge of a balcony \(5.1 \mathrm{~m}\) above the ground. The unit is emitting sound uniformly in all directions. By accident, it falls from rest off the balcony and continues to play on the way down. A gardener is working in a flower bed directly below the falling unit. From the instant the unit begins to fall, how much time is required for the sound intensity level heard by the gardener to increase by \(10.0 \mathrm{~dB} ?\)

Multiple-Concept Example 4 presents one method for modeling this type of problem. Civil engineers use a transit theodolite when surveying. A modern version of this device determines distance by measuring the time required for an ultrasonic pulse to reach a target, reflect from it, and return. Effectively, such a theodolite is calibrated properly when it is programmed with the speed of sound appropriate for the ambient air temperature. (a) Suppose the round-trip time for the pulse is \(0.580 \mathrm{~s}\) on a day when the air temperature is \(293 \mathrm{~K},\) the temperature for which the instrument is calibrated. How far is the target from the theodolite? (b) Assume that air behaves as an ideal gas. If the air temperature were \(298 \mathrm{~K}\), rather than the calibration temperature of \(293 \mathrm{~K}\), what percentage error would there be in the distance measured by the theodolite?

Refer to Interactive Solution \(\underline{16.77}\) at for one approach to this type of problem. Two trucks travel at the same speed. They are far apart on adjacent lanes and approach each other essentially head-on. One driver hears the horn of the other truck at a frequency that is 1.14 times the frequency he hears when the trucks are stationary. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). At what speed is each truck moving?
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