/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A light wire is tightly stretche... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 24

A light wire is tightly stretched with tension F. Transverse traveling waves of amplitude \(A\) and wavelength \(\lambda_1\) carry average power \(P_{av,1} = 0.400\) W. If the wavelength of the waves is doubled, so \(\lambda_2 = 2\lambda_1\), while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{av,2}\) carried by the waves?

Short Answer

Expert verified
The new average power is 0.100 W.

Step by step solution

01

Understand the relation between wavelength and power

For waves stretched along a wire, the average power carried by the waves is related to the tension in the wire, the amplitude of the wave, and the wavelength. The formula for average power \(P_{av}\) is given by:\[P_{av} = \frac{1}{2} \mu v \omega^2 A^2 \]where \(\mu\) is the linear density, \(v\) is the wave speed, \(\omega\) is the angular frequency, and \(A\) is the amplitude.
02

Relate wave speed with tension and wavelength

The wave speed \(v\) on the wire is related to tension \(F\) and linear density \(\mu\) by \(v = \sqrt{\frac{F}{\mu}}\). Furthermore, the wave speed is also related to the wavelength and frequency by \(v = f \lambda\). Since frequency \(f\) can be expressed in terms of angular frequency \(\omega\) as \(f = \frac{\omega}{2\pi}\), substitute back into wave speed, we have:\[v = \frac{\omega \lambda}{2\pi} \]
03

Examine effect of doubling wavelength

If we double the wavelength to \(\lambda_2 = 2\lambda_1\), without changing \(F\), \(A\), or \(\mu\), the wave speed \(v\) remains the same. Thus, the frequency changes as \(f_2 = \frac{f_1}{2}\), and since \(\omega = 2\pi f\), \(\omega_2 = \frac{\omega_1}{2}\).
04

Calculate new power with changed wavelength

Now, substitute into the power expression:\[P_{av,2} = \frac{1}{2} \mu v \left(\frac{\omega_1}{2}\right)^2 A^2 \]Simplify to find the relation between \(P_{av,1}\) and \(P_{av,2}\):\[P_{av,2} = \frac{1}{4} \left(\frac{1}{2} \mu v \omega_1^2 A^2\right) = \frac{1}{4} P_{av,1}\]This shows that when the wavelength is doubled, the average power is reduced to one-fourth.
05

Calculate new average power

Given \(P_{av,1} = 0.400\) W, the new average power is:\[P_{av,2} = \frac{1}{4} \times 0.400 \text{ W} = 0.100 \text{ W}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Wave Speed Calculation
Understanding wave speed is essential when discussing waves on a stretched wire. The speed of a wave, denoted as \( v \), depends on two main factors: the tension \( F \) exerted on the wire and its linear density \( \mu \). This relationship is expressed by the formula:
  • \( v = \sqrt{\frac{F}{\mu}} \)
This equation shows that increasing the tension in the wire will increase the wave speed if the linear density remains unchanged. Conversely, higher linear density will reduce wave speed if the tension stays constant.
This relationship is crucial because wave speed \( v \) helps in determining other wave properties, such as frequency and wavelength. Thus, understanding how tension and linear density affect wave speed is foundational for studying waves.
Wavelength Effect on Power
The wavelength of a wave, represented as \( \lambda \), significantly impacts the power carried by the waves on a wire. Power, specifically average power \( P_{av} \), can be determined using the formula:
  • \( P_{av} = \frac{1}{2} \mu v \omega^2 A^2 \)
Here, \( \omega \) is the angular frequency, \( A \) is the amplitude, and \( \mu \) is the linear density.
When the wavelength is altered, it directly influences the frequency \( f \) and angular frequency \( \omega \) of the wave. If the wavelength doubles, like from \( \lambda_1 \) to \( 2\lambda_1 \), the frequency decreases by half as \( f = \frac{\omega}{2\pi} \). Since power is proportional to the square of the angular frequency \( \omega^2 \), halving the frequency results in a quarter of the power.
This means doubled wavelength results in a significant reduction of power, showcasing the profound effect wavelength has on wave energy.
Tension in a Wire
Tension is a key factor in determining the properties of waves traveling through a wire. The term 'tension' refers to the stretching force applied along the wire. It directly influences the wave speed, critical for analyzing wave phenomena.
  • Formula for wave speed: \( v = \sqrt{\frac{F}{\mu}} \)
Here, \( F \) stands for the tension in the wire, and \( \mu \) represents the linear mass density.
By increasing tension, the wire becomes capable of supporting faster-traveling waves, affecting their frequency and energy. In this exercise, tension remained constant, demonstrating unchanged wave speed despite wavelength changes.
The unchanging tension showcases a stability in the medium, which is essential for consistency in experiments and calculations involving wave mechanics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(Mg\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu\). You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M\); then you measure the node-to-node distance \(d\) for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t) = (8.50 \, \mathrm{mm}) \mathrm{cos}(172 \, \mathrm{rad/m} \space x - 4830 \, \mathrm{rad/s} \space t)$$ Assume that the tension of the string is constant and equal to \(W\). (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W\)? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling \(down\) the string?

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some 200,000 people. Satellites observing these waves from space measured 800 km from one wave crest to the next and a period between waves of 1.0 hour. What was the speed of these waves in m/s and in km/h? Does your answer help you understand why the waves caused such devastation?

A certain transverse wave is described by $$y(x, t) = (6.50 \, \mathrm{mm}) \mathrm{cos} \, 2\pi \Big( \frac{x}{28.0 \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \Big)$$ Determine the wave's (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m long. The density of the rock is 3200 kg/m\(^3\). The mass of the wire is small enough that its effect on the tension in the wire can be ignored. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is 42.0 Hz. When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is 28.0 Hz. What is the density of the liquid?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Translational Dynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fluids

Read Explanation

Energy Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.