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Chapter 16: 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 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.

Short Answer

Expert verified
(a) A is 7.45 dB more powerful than B. (b) A will not sound more than twice as loud as B.

Step by step solution

01

Understand the Formula

The given formula \(\beta=(10 \mathrm{~dB}) \log \left(P / P_{0}\right)\) relates the power levels \(P\) and \(P_0\) for comparing them in decibels. Here, \(\beta\) is the decibel level, \(P\) is the power of amplifier A, and \(P_0\) is the power of amplifier B.
02

Substitute Power Ratings

Substitute the power ratings of amplifier A and B into the formula. For amplifier A, \(P = 250\) watts, and for amplifier B, \(P_0 = 45\) watts. The equation becomes: \[\beta = (10 \mathrm{~dB}) \log \left(\frac{250}{45}\right)\]
03

Calculate the Ratio

First, calculate the ratio of \(P\) to \(P_0\): \(\frac{250}{45} \approx 5.556\).
04

Apply Logarithm and Calculate Decibels

Calculate the logarithm of the ratio from the previous step: \[\log(5.556) \approx 0.745\]. Then, multiply by 10 dB: \[\beta = 10 \cdot 0.745 \approx 7.45\] dB.
05

Step 5a: Answer Part (a)

Amplifier A is approximately 7.45 dB more powerful than amplifier B.
06

Step 5b: Answer Part (b) and Justify

The perception of loudness is logarithmic; a 10 dB increase is perceived as approximately twice as loud, and since 7.45 dB is less than 10 dB, amplifier A will not sound more than twice as loud as amplifier B.

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Key Concepts

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

Power Levels
In the realm of sound and electronics, power levels indicate the amount of energy an amplifier can output. Power, typically measured in watts, describes how much energy is transferred per unit of time.
Power level comparisons help in evaluating the potential loudness of audio equipment like stereo amplifiers. By comparing power levels, you can understand how much more powerful or efficient one device might be compared to another.
For instance, in the problem given, amplifier A has a power of 250 watts per channel, whereas amplifier B is rated at 45 watts per channel. These numbers show the ability of each amplifier to deliver sound; the higher the wattage, ideally, the louder and more robust the sound output could be. Understanding these power levels can guide users in selecting audio equipment appropriate for their venue size and preferred sound experience.
Logarithmic Scale
A logarithmic scale is a way of displaying information that can span a massive range of values compactly and comprehensibly. Unlike a linear scale, where increments grow by addition, a logarithmic scale grows exponentially by multiplication.
In the context of decibels, using a logarithmic scale is essential. This is because human hearing perceives sound intensity in a similar logarithmic manner: changes in perceived loudness correspond to proportional changes, not equal increments.
The decibel scale simplifies comparing vastly different power levels. Instead of dealing with cumbersome mathematical multipliers, the logarithmic formula, \(\beta = 10 \, \text{dB} \cdot \log\left(\frac{P}{P_0}\right)\), allows for a more intuitive grasp by converting the ratio of power levels into the more manageable decibel (dB) units. This transformation sheds light on how much more intense or powerful one scenario is compared to another in terms of sound.
Amplifier Power Comparison
Comparing amplifier power involves understanding how different ratings influence perceived loudness. When the power of two amplifiers is compared using decibels, you can quantify not only their difference in capability but also their impact on perceived sound intensity.
In our situation, amplifier A's power, at 250 watts, is significantly higher than amplifier B's 45 watts. The formula \(\beta = 10 \, \text{dB} \cdot \log\left(\frac{250}{45}\right)\) calculates this difference in decibels, yielding approximately 7.45 dB.
Decibels help in understanding this scale of difference in human sound perception. Although 7.45 dB shows that amplifier A is more powerful, it is not twice as loud as amplifier B. A 10 dB change often signifies a doubling or halving of perceived loudness; hence, 7.45 dB indicates that A is louder, but not by that significant margin.

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Most popular questions from this chapter

The speed of a transverse wave on a string is \(450 \mathrm{~m} / \mathrm{s}\), and the wavelength is \(0.18 \mathrm{~m}\). The amplitude of the wave is \(2.0 \mathrm{~mm}\). How much time is required for a particle of the string to move through a total distance of \(1.0 \mathrm{~km}\) ?

Consult Interactive Solution \(\underline{16.17}\) at in order to review a model for solving this problem. To measure the acceleration due to gravity on a distant planet, an astronaut hangs a \(0.055-\mathrm{kg}\) ball from the end of a wire. The wire has a length of \(0.95 \mathrm{~m}\) and a linear density of \(1.2 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). Using electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of \(0.016 \mathrm{~s}\). The mass of the wire is negligible compared to the mass of the ball. Determine the acceleration due to gravity.

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 \(10.0 \mathrm{~dB} ?\)

Concept Questions The table shows three situations in which the Doppler effect may arise. The first two columns indicate the velocities of the sound source and the observer, where the length of each arrow is proportional to the speed. For each situation, fill in the empty columns by deciding whether the wavelength of the sound and the frequency heard by the observer increase, decrease, or remain the same compared to the case when there is no Doppler effect. Provide a reason for each answer. $$ \begin{array}{|l|c|c|c|c|} \hline & \begin{array}{c} \text { Velocity of Sound } \\ \text { Source (Toward the } \\ \text { Observer) } \end{array} & \begin{array}{c} \text { Velocity of } \\ \text { Observer (Toward } \\ \text { the Source) } \end{array} & \text { Wavelength } & \begin{array}{c} \text { Frequency Heard by } \\ \text { Observer } \end{array} \\ \hline \text { (a) } & 0 \mathrm{~m} / \mathrm{s} & 0 \mathrm{~m} / \mathrm{s} & & \\ \hline \text { (b) } & \rightarrow & 0 \mathrm{~m} / \mathrm{s} & & \\ \hline \text { (c) } & \rightarrow & \leftarrow & & \\ \hline \end{array} $$ Problem The siren on an ambulance is emitting a sound whose frequency is \(2450 \mathrm{~Hz}\). The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). (a) If the ambulance is stationary and you (the "observer") are sitting in a parked car, what is the wavelength of the sound and the frequency heard by you? (b) Suppose the ambulance is moving toward you at a speed of \(26.8 \mathrm{~m} / \mathrm{s}\). Determine the wavelength of the sound and the frequency heard by you. (c) If the ambulance is moving toward you at a speed of \(26.8 \mathrm{~m} / \mathrm{s}\) and you are moving toward it at a speed of \(14.0 \mathrm{~m} / \mathrm{s}\), find the wavelength of the sound and the frequency that you hear. Be sure that your answers are consistent with your answers to the Concept Questions.

A longitudinal wave with a frequency of \(3.0 \mathrm{~Hz}\) takes \(1.7 \mathrm{~s}\) to travel the length of a \(2.5-\mathrm{m}\) Slinky (see Figure \(16-3\) ). Determine the wavelength of the wave.
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