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A horizontal wire is stretched with a tension of 94.0 \(\mathrm{N}\) and the speed of transverse waves for the wire is 492 \(\mathrm{m} / \mathrm{s}\) . What must the amplitude of a traveling wave of frequency 69.0 \(\mathrm{Hz}\) be in order for the average power carried by the wave to be 0.365 \(\mathrm{w}\) ?

Short Answer

Expert verified
The amplitude must be approximately 0.00154 meters.

Step by step solution

01

Understand the Power Formula for a Wave

The average power carried by a wave is given by the formula \( P = \frac{1}{2} \mu A^2 \omega^2 v \), where \( P \) is the power, \( \mu \) is the mass per unit length of the wire, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( v \) is the speed of the wave. Here, we need to find the amplitude \( A \).
02

Determine the Linear Mass Density

The mass per unit length \( \mu \) is related to the tension \( T \) and the speed \( v \) by the equation \( v = \sqrt{\frac{T}{\mu}} \). Solving for \( \mu \), we get \( \mu = \frac{T}{v^2} \). Inserting the values \( T = 94.0 \ \mathrm{N} \) and \( v = 492 \ \mathrm{m/s} \), we calculate \( \mu = \frac{94.0}{492^2} \).
03

Calculate the Angular Frequency

The angular frequency \( \omega \) is given by \( \omega = 2\pi f \), where \( f \) is the frequency. Given \( f = 69.0 \ \mathrm{Hz} \), we find that \( \omega = 2\pi \times 69.0 \).
04

Solve for the Amplitude

We rearrange the power equation \( P = \frac{1}{2} \mu A^2 \omega^2 v \) to solve for \( A^2 \): \( A^2 = \frac{2P}{\mu \omega^2 v} \). Substituting the known values \( P = 0.365 \), \( \mu \), \( \omega \), and \( v = 492 \), we can solve for \( A^2 \), and then take the square root to find \( A \).
05

Calculation and Result

Calculate \( \mu = \frac{94.0}{492^2} \approx 0.000389\) \( \mathrm{kg/m} \). Next, \( \omega = 2\pi \times 69.0 \approx 433.54 \ \mathrm{rad/s} \). Substituting these into the formula for \( A^2 \), \( A^2 = \frac{2 \times 0.365}{0.000389 \times (433.54)^2 \times 492} \approx 2.37 \times 10^{-6} \). Finally, \( A = \sqrt{2.37 \times 10^{-6}} \approx 0.00154 \ \mathrm{m} \).

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Key Concepts

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

average power
When studying waves, understanding average power is essential as it describes the energy transmitted through a medium over time. The average power of a wave is determined by several factors including the wave's amplitude, frequency, speed, and the mass per unit length of the medium.
The formula to calculate the average power, \( P \), for a wave travelling through a wire is given by:\[ P = \frac{1}{2} \mu A^2 \omega^2 v \]where:
  • \( \mu \) is the mass per unit length
  • \( A \) is the amplitude of the wave
  • \( \omega \) is the angular frequency
  • \( v \) is the speed of the wave

In this equation, you can see that the amplitude has a squared relationship with the power, meaning that a small change in amplitude can have a large effect on the power. Therefore, understanding how each of these elements affect power is crucial when exploring wave dynamics.
transverse waves
Transverse waves are a type of wave where the movement of the medium is perpendicular to the direction of wave propagation. In simpler terms, if the wave travels horizontally along a wire, the particles of the medium (the wire) move up and down.
Because the particles move in a direction perpendicular to the direction of energy transfer, transverse waves often generate a ripple effect. This is seen in water waves or vibrations traveling through a guitar string.
In physics problems, like the one we are dealing with, understanding the nature of transverse waves is important for calculating properties like speed, tension, and wavelength. Unlike longitudinal waves, where the particles move in the same direction as the wave, transverse waves offer a visual representation of wave phenomena.
mass per unit length
Mass per unit length, often denoted as \( \mu \), plays a crucial role in determining the speed and behavior of waves along a medium. It's simply the mass of the medium divided by its length, providing an idea of how much matter is distributed along a section of the medium.
For this particular exercise involving a wire under tension, the mass per unit length affects the speed of waves traveling through the wire, which is calculated using the equation:\[v = \sqrt{\frac{T}{\mu}} \]where:
  • \( T \) is the tension in the wire
  • \( \mu \) is the mass per unit length
This highlights that the wave's speed is inversely proportional to the mass per unit length. Hence, a heavier wire (with higher \( \mu \)) under the same tension would result in slower wave speeds. By calculating \( \mu \) accurately, you ensure precision in solving wave-related problems.
angular frequency
Angular frequency, denoted as \( \omega \), is a measure of how quickly a wave oscillates in radians per second. It is closely related to the regular frequency \( f \), which is measured in cycles per second (Hz), by the formula:\[ \omega = 2\pi f\]The constant \( 2\pi \) converts cycles to radians, giving a sense of how many radians a wave phase advances per second.
For instance, if a wave has a frequency of \( 69\, \mathrm{Hz} \), the angular frequency would be:\[ \omega = 2\pi \times 69\]Knowing the angular frequency helps in understanding the oscillatory behavior of waves and is crucial when calculating average power, particularly since it appears squared in the power formula. The relationship between frequency and angular frequency simplifies the mathematical analysis of oscillating systems, making it easier to predict wave behaviors over time.

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Most popular questions from this chapter

A deep-sea diver is suspended beneath the surface of Loch Ness by a \(100-\) m-long cable that is attached to a boat on the surface (Fig. P15.84). The diver and his suit have a total mass of 120 \(\mathrm{kg}\) and a volume of 0.0880 \(\mathrm{m}^{3} .\) The cable has a diameter of 2.00 \(\mathrm{cm}\) and a linear mass density of \(\mu=\) 1.10 \(\mathrm{kg} / \mathrm{m} .\) The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat. (a) What is the tension in the cable at its lower end, where it is attached to the diver? Do not forget to include the buoyant force that the water (density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) ) exerts on him. (b)Calculate the tension in the cable a distance \(x\) above the diver. The buoyant force on the cable must be included in your calculation. (c) The speed of transverse waves on the cable is given by \(v=\sqrt{F / \mu}\) (Eq. 15.13). The speed therefore varies along the cable, since the tension is not constant. (This expression neglects the damping force that the water exerts on the moving cable.. Integrate to find the time required for the first signal to reach the surface.

Combining Standing Waves. A guitar string of length \(L\) is plucked in such a way that the total wave produced is the sum of the fundamental and the second harmonic. That is, the standing wave is given by $$y(x, t)=y_{1}(x, t)+y_{2}(x, t)$$ where $$\begin{aligned} y_{1}(x, t) &=C \sin \omega_{1} t \sin k_{1} x \\ y_{2}(x, t) &=C \sin \omega_{2} t \sin k_{2} x \end{aligned}$$ with \(\omega_{1}=v k_{1}\) and \(\omega_{2}=v k_{2}\) . (a) At what values of \(x\) are the nodes of \(y_{1} ?\left(\) b) At what values of \(x\) are the nodes of \(y_{2} ?(\mathrm{c})\) Graph \right the total wave at \(t=0, t=\frac{1}{8} f_{1}, t=\frac{1}{4} f_{1}, t=\frac{3}{8} f_{1},\) and \(t=\frac{1}{2} f_{1}\) . (d) Does the sum of the two standing waves \(y_{1}\) and \(y_{2}\) produce a standing wave? Explain.

A thin, 75.0 -cm wire has a mass of 16.5 \(\mathrm{g} .\) One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 \(\mathrm{cm}\) makes 875 vibrations per second? (b) How fast would this wave travel?

A sinusoidal transverse wave travels on a string. The string has length 8.00 \(\mathrm{m}\) and mass 6.00 \(\mathrm{g} .\) The wave speed is \(30.0 \mathrm{m} / \mathrm{s},\) and the wavelength is 0.200 \(\mathrm{m} .\) (a) If the wave is to have an average power of \(50.0 \mathrm{W},\) what must be the amplitude of the wave? (b) For this same string, if the amplitude and wave-length are the same as in part (a), what is the average power for the wave if the tension is increased such the wave speed is doubled?

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 m from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

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