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Chapter 15: Problem 21

(II) (a) Show that the average rate with which energy is transported along a cord by a mechanical wave of frequency \(f\) and amplitude \(A\) is $$\bar{P}=2 \pi^{2} \mu v f^{2} A^{2}$$ where \(v\) is the speed of the wave and \(\mu\) is the mass per unit length of the cord. (b) If the cord is under a tension \(F_{\mathrm{T}}=135 \mathrm{~N}\) and has mass per unit length \(0.10 \mathrm{~kg} / \mathrm{m},\) what power is required to transmit \(120-\mathrm{Hz}\) transverse waves of amplitude \(2.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
The average power required is approximately 266 W.

Step by step solution

01

Understanding Wave Properties

We are given a mechanical wave traveling along a cord. The parameters include frequency \(f\), amplitude \(A\), wave speed \(v\), and mass per unit length \(\mu\). The task is to show that the average power \(\bar{P}\) is given by the formula \(\bar{P}=2 \pi^{2} \mu v f^{2} A^{2}\).
02

Deriving the Power Equation

The power transported by a wave is calculated based on the wave speed and energy density. The energy density for a wave on a string is related to amplitude and frequency, leading to power \(\bar{P} = \frac{1}{2} \mu v (2 \pi f A)^{2}\). Simplifying this expression, we use the identity \((2\pi f A)^{2} = 4\pi^2 f^2 A^2\) to obtain \(\bar{P}=2 \pi^{2} \mu v f^{2} A^{2}\).
03

Calculate Wave Speed

The wave speed \(v\) on a string is given by \(v = \sqrt{\frac{F_{T}}{\mu}}\), where \(F_{T} = 135 \text{ N} \) is the tension and \(\mu = 0.10 \text{ kg/m}\) is mass per unit length. Thus, \(v = \sqrt{\frac{135}{0.10}} = 36.7 \text{ m/s}\).
04

Substituting Values Into Power Equation

Substitute \(f = 120 \text{ Hz}\), \(A = 2.0 \times 10^{-2} \text{ m}\), \(\mu = 0.10 \text{ kg/m}\), and \(v = 36.7 \text{ m/s}\) into the power formula \(\bar{P}=2 \pi^{2} \mu v f^{2} A^{2}\). This yields \(\bar{P}=2 \pi^{2} (0.10) (36.7) (120)^{2} (2.0 \times 10^{-2})^{2}\).
05

Solving for Power Required

Calculate \(\bar{P} = 2 \times \pi^{2} \times 0.1 \times 36.7 \times 120^{2} \times (2.0 \times 10^{-2})^{2} \). Solving this gives a power of approximately \(266 \text{ W}\).

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Key Concepts

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

Energy Transport
When we talk about energy transport in mechanical waves, we're essentially discussing how energy is moved from one point to another through a medium. For waves on a string, this energy transport is characterized by the average power \( \bar{P} \). The expression \( \bar{P}=2 \pi^{2} \mu v f^{2} A^{2}\) helps us understand how energy transfer depends on various factors like wave speed, frequency, amplitude, and the mass per unit length of the string.
  • **Wave Speed (\(v\))**: Faster waves carry energy more quickly.
  • **Frequency (\(f\))**: Higher frequency means energy is transferred faster.
  • **Amplitude (\(A\))**: Larger amplitude leads to greater energy transfer.
  • **Mass per Unit Length (\(\mu\))**: Denser strings impact energy transport efficiency.
These components work together to determine how efficiently energy is transferred along the wave, making \( \bar{P} \) a crucial concept in wave dynamics.
Wave Speed
Wave speed is a key variable in wave mechanics. For waves on a string, the speed \(v\) is determined by the tension in the string \(F_{T}\) and the mass per unit length \(\mu\). The relationship is given by the formula \(v=\sqrt{\frac{F_{T}}{\mu}}\).
  • **Tension (\(F_{T}\))**: The force pulling on the string, usually measured in Newtons.
  • **Mass per Unit Length (\(\mu\))**: It's like the 'density' of the string, in terms of mass, and impacts how fast the wave travels.
To increase the wave speed, you could either increase the tension or decrease the mass per unit length. For instance, in the given problem, by calculating \(v=\sqrt{\frac{135}{0.10}}\), the speed of the wave is found to be approximately \(36.7 \text{ m/s}\). This shows the practical interplay between force applied and string characteristics in determining wave behavior.
Amplitude
Amplitude is the maximum displacement of particles in the medium from their rest position as the wave passes through. It is a measure of the wave's intensity or strength -- in this case, how high or low the peaks of the wave are.
  • **Measurement**: Amplitude is often measured in meters (m), especially in mechanical waves on strings.
  • **Impact on Energy**: Amplitude is crucial because in the power formula \( \bar{P}=2 \pi^{2} \mu v f^{2} A^{2} \), the power is directly proportional to the square of the amplitude \((A^2)\). Thus, doubling the amplitude will quadruple the power transmitted.
In the problem, an amplitude of \(2.0 \text{ cm}\) (or \(0.02 \text{ m}\)) is significant in calculating the power needed to transmit the wave. Amplitude serves as a reminder of how much energy each wave cycle can transport.
Mass per Unit Length
Mass per unit length (\(\mu\)) is a critical characteristic of the medium through which the wave travels. It describes how much mass exists in a given length of the string or cord on which the wave is propagating.
  • **Understanding **: Measured in kilograms per meter (kg/m), it is a way to express the linear density of the string.
  • **Role in Wave Dynamics**: A higher mass per unit length means the wave will travel more slowly if tension remains constant, as evidenced by the wave speed formula \(v = \sqrt{\frac{F_{T}}{\mu}}\).
In the given problem, \(\mu = 0.10 \text{ kg/m}\) represents the string's density. This value affects both the wave speed and the energy needed to transport waves along the string. Lower density strings mean less energy is needed for the same wave parameters, while higher mass per unit length demands more energy due to the increased inertia.

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Most popular questions from this chapter

(II) One end of a horizontal string of linear density \(6.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\) is attached to a small-amplitude mechan ical \(120-\mathrm{Hz}\) oscillator. The string passes over a pulley, a distance \(\ell=1.50 \mathrm{~m}\) away, and weights are hung from this end, Fig. \(15-37\). What mass \(m\) must be hung from this end of the string to produce \((a)\) one loop, (b) two loops, and (c) five loops of a standing wave? Assume the string at the oscillator is a node, which is nearly true.

(II) \(\mathrm{A} .0 .40\) -kg cord is stretched between two supports, 7.8 \(\mathrm{m}\) apart. When one support is struck by a hammer, a transverse wave travels down the cord and reaches the other support in 0.85 \(\mathrm{s}\) . What is the tension in the cord?

(II) The displacement of a standing wave on a string is given by \(D=2.4 \sin (0.60 x) \cos (42 t),\) where \(x\) and \(D\) are in centimeters and \(t\) is in seconds. (a) What is the distance (cm) between nodes? (b) Give the amplitude, frequency, and speed of each of the component waves. (c) Find the speed of a particle of the string at \(x=3.20 \mathrm{~cm}\) when \(t=2.5 \mathrm{~s}\).

(I) A fisherman notices that wave crests pass the bow of his anchored boat every \(3.0 \mathrm{~s}\). He measures the distance between two crests to be \(8.0 \mathrm{~m}\). How fast are the waves traveling?

(II) A standing wave on a 1.64 -m-long horizontal string displays three loops when the string vibrates at \(120 \mathrm{~Hz}\). The maximum swing of the string (top to bottom) at the center of each loop is \(8.00 \mathrm{~cm} .\) ( \(a\) ) What is the function describing the standing wave? (b) What are the functions describing the two equal-amplitude waves traveling in opposite directions that make up the standing wave?
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