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Average Power A string along which waves can travel is \(2.70 \mathrm{~m}\) long and has a mass of \(260 \mathrm{~g}\). The tension in the string is \(36.0 \mathrm{~N}\). What must be the frequency of traveling waves of amplitude \(7.70\) \(\mathrm{mm}\) for the average power to be \(85.0 \mathrm{~W} ?\)

Short Answer

Expert verified
The frequency must be 166.8 Hz.

Step by step solution

01

- Find the Linear Density of the String

The linear density \( \mu \) of the string is given by the mass per unit length. Convert the mass from grams to kilograms for consistency in SI units and use the formula \( \mu = \frac{m}{L} \). Substitute \( m = 0.260 \) kg and \( L = 2.70 \) m: \[ \mu = \frac{0.260 \text{ kg}}{2.70 \text{ m}} = 0.0963 \text{ kg/m} \]
02

- Calculate the Wave Speed

The wave speed \( v \) on a string under tension \( T \) is given by \( v = \sqrt{\frac{T}{\mu}} \). Substitute \( T = 36.0 \) N and \( \mu = 0.0963 \text{ kg/m} \): \[ v = \sqrt{\frac{36.0 \text{ N}}{0.0963 \text{ kg/m}}} = 19.3 \text{ m/s} \]
03

- Use the Average Power Formula

For a sinusoidal wave, the average power \( P_{avg} \) is given by \[ P_{avg} = \frac{1}{2} \mu \omega^2 A^2 v \] \Rearrange the formula to solve for the angular frequency \( \omega \): \[ \omega = \sqrt{ \frac{2 P_{avg}}{ \mu A^2 v }} \] Substitute \( P_{avg} = 85.0 \) W, \( \mu = 0.0963 \text{ kg/m} \), \( A = 7.70 \) mm \( = 7.70 \times 10^{-3} \) m, and \( v = 19.3 \text{ m/s} \): \[ \omega = \sqrt{ \frac{2 \times 85.0 \text{ W}}{0.0963 \text{ kg/m} \times (7.70 \times 10^{-3} \text{ m})^2 \times 19.3 \text{ m/s}}} = 1047.2 \text{ rad/s} \]
04

- Convert Angular Frequency to Frequency

Frequency \( f \) is related to angular frequency \( \omega \) by \( f = \frac{\omega}{2\pi} \). \Substitute \( \omega = 1047.2 \text{ rad/s} \): \[ f = \frac{1047.2 \text{ rad/s}}{2\pi} = 166.8 \text{ Hz} \]

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Key Concepts

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

Linear Density
Linear density, often symbolized as \( \mu \), is a measure of mass per unit length. It is fundamental in wave mechanics, especially for waves traveling along strings or ropes. Calculating linear density involves dividing the mass of an object by its length. For instance, if we have a string that is 2.70 meters long and has a mass of 260 grams, we first convert the mass to kilograms (0.260 kg), then use the formula:
\( \mu = \frac{m}{L} = \frac{0.260 \text{ kg}}{2.70 \text{ m}} = 0.0963 \text{ kg/m} \).
This calculation sets the stage for understanding wave speed and related concepts.
Wave Speed
Wave speed, denoted by \( v \), describes how fast a wave travels through a medium. For a string under tension, the wave speed depends on both the tension and the linear density of the string. This is expressed with the formula:
\( v = \sqrt{\frac{T}{\mu}} \)
where \( T \) is the tension and \( \mu \) is the linear density. For a string with tension 36.0 N and linear density 0.0963 kg/m, we can find the wave speed by calculating:
\( v = \sqrt{\frac{36.0 \text{ N}}{0.0963 \text{ kg/m}}} = 19.3 \text{ m/s} \).
Understanding wave speed is essential for determining other wave properties such as frequency and power.
Angular Frequency
Angular frequency, represented by \( \omega \), measures how rapidly the wave oscillates in radians per second. It is related to the frequency of the wave but gives a different perspective. Angular frequency can be found using the average power formula for sinusoidal waves:
\[ P_{avg} = \frac{1}{2} \mu \omega^2 A^2 v \]
Rearranging to solve for \( \omega \), we get:
\( \omega = \sqrt{ \frac{2 P_{avg}}{ \mu A^2 v }} \). Substituting known values (\( P_{avg} = 85.0 \text{ W} \), \( \mu = 0.0963 \text{ kg/m} \), \( A = 7.70 \times 10^{-3} \text{ m} \), and \( v = 19.3 \text{ m/s} \)):
\( \omega = \sqrt{ \frac{2 \times 85.0 \text{ W}}{0.0963 \text{ kg/m} \times (7.70 \times 10^{-3} \text{ m})^2 \times 19.3 \text{ m/s}}} = 1047.2 \text{ rad/s} \).
This concept bridges frequency and energy in wave dynamics.
Average Power Formula
The average power of a wave traveling along a string indicates how much energy is transported per second. For sinusoidal waves, the average power \( P_{avg} \) is given by:
\[ P_{avg} = \frac{1}{2} \mu \omega^2 A^2 v \]
This formula encompasses several key wave properties: linear density \( \mu \), angular frequency \( \omega \), amplitude \( A \), and wave speed \( v \). Each term plays a crucial role:
  • \( \mu \): dictates mass effect
  • \( \omega \): relates to oscillation speed
  • \( A \): represents wave height
  • \( v \): measures how quickly the wave propagates
Rearranging the formula allows solving for unknown quantities like angular frequency or average power.
Frequency Calculation
Frequency, denoted by \( f \), represents the number of oscillations per second. To relate angular frequency \( \omega \) to frequency, use the formula:
\( f = \frac{\omega}{2\pi} \)
Given \( \omega = 1047.2 \text{ rad/s} \), the frequency is calculated as:
\( f = \frac{1047.2 \text{ rad/s}}{2\pi} \approx 166.8 \text{ Hz} \).
This conversion is essential for understanding how often waves pass through a point per second, thereby giving a complete picture of wave dynamics.

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Most popular questions from this chapter

Three Sinusoidal Waves Three sinusoidal waves of the same frequency travel along a string in the positive direction of an \(x\) axis. Their amplitudes are \(y_{1}, y_{1} / 2\), and \(y_{1} / 3\), and their initial phases are \(0, \pi / 2\), and \(\pi \mathrm{rad}\), respectively. What are (a) the amplitude and (b) the phase constant of the resultant wave? (c) Plot the wave form of the resultant wave at \(t=0.00 \mathrm{~s}\), and discuss its behavior as \(t\) increases.

Write the Equation Write the equation for a sinusoidal wave traveling in the negative direction along an \(x\) axis and having an amplitude of \(0.010 \mathrm{~m}\), a frequency of \(550 \mathrm{~Hz}\), and a speed of \(330 \mathrm{~m} / \mathrm{s}\).

Three-Loop Standing Wave A string \(3.0 \mathrm{~m}\) long is oscillating as a three-loop standing wave with an amplitude of \(1.0 \mathrm{~cm}\). The wave speed is \(100 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency? (b) Write equations for two waves that, when combined, will result in this standing wave.

Uniform Rope A uniform rope of mass \(m\) and length \(L\) hangs from a ceiling. (a) Show that the speed of a transverse wave on the rope is a function of \(y\), the distance from the lower end, and is given by \(v^{\text {wave }}=\sqrt{g y}\). (b) Show that the time a transverse wave takes to travel the length of the rope is given by \(t=2 \sqrt{L / g}\)

Two Identical Traveling Waves Two identical traveling waves, moving in the same direction, are out of phase by \(\pi / 2\) rad. What is the amplitude of the resultant wave in terms of the common amplitude \(Y\) of the combining waves?

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