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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 wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?

Short Answer

Expert verified
The amplitude of the wave must be \( 0.0641 \, \text{m} \) for an average power of \( 50.0 \, \text{W} \). If the wave speed is doubled, the average power of the wave will be \( 200.0 \, \text{W} \)

Step by step solution

01

Calculate the linear density

The linear mass density \( \mu \) is the mass per unit length. It can be calculated as \( \mu = \frac{M}{L} \), where \( M \) is the mass and \( L \) is the length. Substituting the given values, we get \( \mu = \frac{6.00 \times 10^{-3} \, \text{kg}}{8.00 \, \text{m}} = 0.75 \times 10^{-3} \, \text{kg/m} \)
02

Calculate the angular frequency

The angular frequency \( \omega \) is given by \( 2\pi v/\lambda \), where \( v \) is the wave speed and \( \lambda \) is the wavelength. Substituting the given values, we get \( \omega = \frac{2\pi \times 30.0 \, \text{m/s}}{0.200 \, \text{m}} = 300\pi \, \text{rad/s} \)
03

Calculate the amplitude

Substitute the calculated values of \( \mu \), \( v \), and \( \omega \) in the equation for \( \mathrm{P_{avg}} \), and solve for \( A \). This yields \( A = \sqrt{\frac{2 \times 50.0 \, \text{W}}{0.75 \times 10^{-3} \, \text{kg/m} \times 30.0 \, \text{m/s} \times (300\pi \, \text{rad/s})^2}} = 0.0641 \, \text{m} \)
04

Calculate the power with doubled wave speed

In the second part, it is given that the tension is increased such that the wave speed is doubled. This means the new wave speed is \( 2 \times 30.0 \, \text{m/s} = 60.0 \, \text{m/s} \). The angular frequency with the new wave speed is \( \omega' = 2\pi \times v'/\lambda = 2\pi \times 60.0 \, \text{m/s}/0.200 \, \text{m} = 600\pi \, \text{rad/s} \). Now we substitute these values into the average power equation to find the new power: \( \mathrm{P_{avg}'} = \frac{1}{2}\mu {v'}^2{\omega'}^2A^2 = \frac{1}{2} \times 0.75 \times 10^{-3} \, \text{kg/m} \times (60.0 \, \text{m/s})^2 \times (600\pi \, \text{rad/s})^2 \times (0.0641 \, \text{m})^2 = 200.0 \, \text{W} \)

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Key Concepts

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

Transverse Waves
Transverse waves are a type of wave where the particles of the medium move perpendicular to the direction of the wave propagation. Imagine a string being flicked up and down; the wave travels along the string while the particles of the string move up and down.
Transverse waves are common in many scenarios:
  • Waves on a string, like those during musical performances.
  • Ripples on the surface of the water.
  • Light waves, as they consist of oscillating electric and magnetic fields perpendicular to each other.
In this specific exercise, the wave is traveling along a string, making it a classic example of a sinusoidal transverse wave. The key is understanding that as the wave travels horizontally, the string's displacement is vertical.
Linear Mass Density
Linear mass density (\(\mu\)) is an essential concept when dealing with waves on a string. It tells us how much mass is there per unit length of the string. This property impacts how waves move along the string.

The formula for linear mass density is:\[\mu = \frac{M}{L}\]where \(M\) is the mass and \(L\) is the length of the string. In the problem, we calculated it as:\[\mu = \frac{6.00 \times 10^{-3} \, \text{kg}}{8.00 \, \text{m}} = 0.75 \times 10^{-3} \, \text{kg/m}\]The importance of knowing the linear mass density is that it contributes directly to how wave speed (\(v\)) is calculated, influencing the wave's behavior on the string.
Wave Speed
Wave speed (\(v\)) measures how fast a wave propagates through a medium. For waves on a string, the wave speed depends on the string's tension and its linear mass density. In the basic form, wave speed can be expressed as:\[v = \sqrt{\frac{T}{\mu}}\]where \(T\) is the tension in the string. In the problem, however, the wave speed is provided at 30.0 m/s initially.

The wave speed affects many aspects of wave behavior:
  • The frequency and wavelength, linked by the formula \(v = \lambda f\).
  • The average power carried by the wave.
Doubling the wave speed by increasing the tension significantly impacts these characteristics and must be considered when calculating energy parameters.
Angular Frequency
Angular frequency (\(\omega\)) describes how many rotations occur in a unit of time, often in radians per second. It is crucial in wave mechanics because it determines how quickly the wave oscillates. In the context of the problem, the angular frequency is linked with the wave speed and wavelength by:\[\omega = \frac{2\pi v}{\lambda}\]where:
  • \(v\) is the wave speed
  • \(\lambda\) is the wavelength
For instance, using the given values, the initial angular frequency is calculated as:\[\omega = \frac{2\pi \times 30.0 \, \text{m/s}}{0.200 \, \text{m}} = 300\pi \, \text{rad/s}\]Thus, angular frequency is vital for understanding the vibrational characteristics of the wave, especially when examining how the wave's power changes with different speeds.

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Most popular questions from this chapter

A rectangular neoprene sheet has width \(W=1.00 \mathrm{~m}\) and length \(L=4.00 \mathrm{~m}\). The two shorter edges are affixed to rigid steel bars that are used to stretch the sheet taut and horizontal. The force applied to either end of the sheet is \(F=81.0 \mathrm{~N}\). The sheet has a total mass \(M=4.00 \mathrm{~kg} .\) The left edge of the sheet is wiggled vertically in a uniform sinusoidal motion with amplitude \(A=10.0 \mathrm{~cm}\) and frequency \(f=1.00 \mathrm{~Hz}\). This sends waves spanning the width of the sheet rippling from left to right. The right side of the sheet moves upward and downward freely as these waves complete their traversal. (a) Use a twodimensional generalization of the discussion in Section 15.4 to derive an expression for the velocity with which the waves move along the sheet in terms of generic values of \(W, L, F, M, f,\) and \(A .\) What is the value of this speed for the specified choices of these parameters? (b) If the positive \(x\) -axis is oriented rightward and the steel bars are parallel to the \(y\) -axis, the height of the sheet may be characterized as \(z(x, y)=A \sin (k x-\omega t)\) What is the value of the wave number \(k ?\) (c) Write down an expression with generic parameters for the rate of rightward energy transfer by the slice of sheet at a given value of \(x\) at generic time \(t\). (d) The power at \(x=0\) is supplied by the agent wiggling the left bar upward and downward. How much energy is supplied each second by that agent? Express your answer in terms of generic parameters and also as a specific energy for the given parameters.

One string of a certain musical instrument is \(75.0 \mathrm{~cm}\) long and has a mass of \(8.75 \mathrm{~g}\). It is being played in a room where the speed of sound is \(344 \mathrm{~m} / \mathrm{s}\). (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength \(0.765 \mathrm{~m} ?\) (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

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