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Chapter 15: Problem 22

A piano wire with mass \(3.00 \mathrm{~g}\) and length \(80.0 \mathrm{~cm}\) is stretched with a tension of \(25.0 \mathrm{~N}\). A wave with frequency \(120.0 \mathrm{~Hz}\) and amplitude \(1.6 \mathrm{~mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

Short Answer

Expert verified
The average power carried by the wave with the initially provided amplitude is 0.0485 W. If the amplitude of the wave is halved, the average power reduces to 0.0121 W.

Step by step solution

01

Calculate the linear mass density

First, calculate the linear mass density \( \mu \) using the formula \( \mu = \frac{m}{L} \), where \( m = 3.00 \mathrm{~g} = 0.003 \mathrm{~kg} \) (converting grams to kilograms) and \( L = 80.0 \mathrm{~cm} = 0.8 \mathrm{~m} \) (converting cm to m). Substituting these values into the formula we get \( \mu = \frac{0.003}{0.8} = 0.00375 \mathrm{~kg/m} \).
02

Find the wave speed

Next, calculate the wave speed using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T = 25.0 \mathrm{~N} \). Substituting the given tension and the linear mass density calculated in the previous step, we get \( v = \sqrt{\frac{25}{0.00375}} = 287.228 \mathrm{~m/s} \).
03

Calculate the power carried by the wave

Use the formula \( P = 2 \pi^2 f^2 A^2 v \mu \) to calculate the power. Where \( f = 120 \mathrm{~Hz} \), \( A = 1.6 \mathrm{~mm} = 0.0016 \mathrm{~m} \), \( v = 287.228 \mathrm{~m/s} \) and \( \mu = 0.00375 \mathrm{~kg/m} \). Substituting these values, we get \( P = 2 * \pi^2 * (120)^2 * (0.0016)^2 * 287.228 * 0.00375 = 0.0485 \mathrm{~W} \).
04

Determine the average power when amplitude is halved

By halving the amplitude, the new amplitude \( A' = \frac{A}{2} = 0.0008 \mathrm{~m} \). Substituting this new value into the power formula we get, \( P' = 2 * \pi^2 * (120)^2 * (0.0008)^2 * 287.228 * 0.00375 = 0.0121 \mathrm{~W} \).

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Key Concepts

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

Linear Mass Density
The concept of linear mass density is crucial when dealing with waves on strings, such as those on a piano wire. It's denoted by the Greek letter 碌 and is defined as the mass per unit length of the wire. The formula to calculate it is simply \( \mu = \frac{m}{L} \), where m represents the mass of the wire and L is its length.

Understanding this characteristic is the first step in determining how waves behave on the wire. Because the tension in the wire and the linear mass density influence wave speed, this property also indirectly affects the power transmitted by a wave. In our exercise, we used the wire's mass and length to calculate this value, which appeared to be \(0.00375 \mathrm{~kg/m}\).
Wave Speed
The speed at which waves travel through a medium, such as a piano wire, is known as wave speed. For a string under tension, the wave speed can be calculated with the equation \( v = \sqrt{\frac{T}{\mu}} \), with T representing the tension in the string and 碌 being the linear mass density we previously discussed.

Wave speed is intricate, as it's affected by both the medium's properties and the force applied to it. In our calculation, substituting the known tension and the linear mass density gave us a wave speed of \(287.228 \mathrm{~m/s}\), indicating how quickly the energy would propagate along the wire.
Power of Mechanical Waves
In the context of mechanical waves on a string, such as a piano wire, power refers to the rate at which energy is transferred by the wave. It can be expressed through the formula \(P = 2 \pi^2 f^2 A^2 v \mu\), combining wave frequency, amplitude, speed, and linear mass density.

When a wave oscillates on a string, energy is transmitted along it, doing work on the surrounding particles. The power calculation tells us how efficient this energy transfer is. In our problem, we determined the wave's power as \(0.0485 \mathrm{~W}\) by applying the given wave properties.
Wave Frequency
Wave frequency is the number of oscillations a wave undergoes per unit of time, measured in Hertz (Hz). It's intimately connected to both the energy and the power of a wave: the higher the frequency, the more cycles per second, and generally, the more energy the wave carries.

In our exercise, the frequency is given as \(120.0 \mathrm{~Hz}\), which signifies that the wave vibrates 120 times each second. This high frequency would typically mean more energy is transported by the wave, and this is directly factored into the power calculation formula we used.
Wave Amplitude
The term wave amplitude describes the maximum displacement of a point on a wave, relative to its rest position. It's a measure of the wave's strength or intensity. For a wave on a string, amplitude relates to how far the string moves from its rest position during an oscillation.

In physics, wave power is strongly dependent on amplitude. When we halved the amplitude in our problem, the power transmitted by the wave didn't just decrease incrementally, it dropped to a quarter of its original value, since power depends on the square of the amplitude. This drastic change, from \(0.0485 \mathrm{~W}\) to \(0.0121 \mathrm{~W}\), shows the importance of amplitude in wave energy transmission.

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Most popular questions from this chapter

You are designing a two-string instrument with metal strings \(35.0 \mathrm{~cm}\) long, as shown in Fig. \(\mathrm{P} 15.52 .\) Both strings are under the same tension. String \(S_{1}\) has a mass of \(8.00 \mathrm{~g}\) and produces the note middle \(\mathrm{C}\) (frequency \(262 \mathrm{~Hz}\) ) in its fundamental mode. (a) What should be the tension in the string? (b) What should be the mass of string \(S_{2}\) so that it will produce A-sharp (frequency \(466 \mathrm{~Hz}\) ) as its fundamental? (c) To extend the range of your instrument, you include a fret located just under the strings but not normally touching them. How far from the upper end should you put this fret so that when you press \(S_{1}\) tightly against it, this string will produce \(\mathrm{C}\) -sharp (frequency \(277 \mathrm{~Hz}\) ) in its fundamental? That is, what is \(x\) in the figure? (d) If you press \(S_{2}\) against the fret, what frequency of sound will it produce in its fundamental?

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is \(0.026 \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(4.3 \mathrm{~m}\) from the source. (a) What is the intensity at a distance of \(3.1 \mathrm{~m}\) from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

For a string stretched between two supports, two successive standing-wave frequencies are \(525 \mathrm{~Hz}\) and \(630 \mathrm{~Hz}\). There are other standing-wave frequencies lower than \(525 \mathrm{~Hz}\) and higher than \(630 \mathrm{~Hz}\). If the speed of transverse waves on the string is \(384 \mathrm{~m} / \mathrm{s},\) what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.

The wave function of a standing wave is \(y(x, t)=(4.44 \mathrm{~mm})\) \(\sin [(32.5 \mathrm{rad} / \mathrm{m}) x] \sin [(754 \mathrm{rad} / \mathrm{s}) t] .\) For the two traveling waves that make up this standing wave, find the (a) amplitude; (b) wavelength; (c) frequency; (d) wave speed; (e) wave functions. (f) From the information given, can you determine which harmonic this is? Explain.

\(\mathrm{A}\) fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is \(y(x, t)=(2.30 \mathrm{~mm}) \cos [(6.98 \mathrm{rad} / \mathrm{m}) x+(742 \mathrm{rad} / \mathrm{s}) t] .\) Being more practical, you measure the rope to have a length of \(1.35 \mathrm{~m}\) and a mass of \(0.00338 \mathrm{~kg}\). You are then asked to determine the following: (a) amplitude; (b) frequency; (c) wavelength; (d) wave speed; (e) direction the wave is traveling; (f) tension in the rope; (g) average power transmitted by the wave.
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