/*! 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 83 A uniform rope with length \(L\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 83

A uniform rope with length \(L\) and mass \(m\) is held at one end and whirled in a horizontal circle with angular velocity \(\omega\) . You can ignore the force of gravity on the rope. Find the time required for a transverse wave to travel from one end of the rope to the other.

Short Answer

Expert verified
The time required for the wave to travel across the rope is \( \frac{1}{\omega} \).

Step by step solution

01

Understand the Wave Speed Formula

The speed of a wave on a rope is given by the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the rope and \( \mu \) is the linear mass density of the rope (mass per unit length). Here, \( \mu = \frac{m}{L} \).
02

Determine the Tension

Since the rope is being whirled in a horizontal circle with angular velocity \( \omega \), the tension in the rope (especially at the endpoint) is due to the centripetal force. The tension \( T \) at a point can be expressed as \( T = m_{segment} \times \omega^2 \times r \). At the end of the rope, \( m_{segment} = m \) and \( r = L \), thus \( T = m \cdot \omega^2 \cdot L \).
03

Calculate the Wave Speed

Substitute the tension \( T = m \cdot \omega^2 \cdot L \) and linear mass density \( \mu = \frac{m}{L} \) into the wave speed formula: \( v = \sqrt{\frac{m \cdot \omega^2 \cdot L}{m/L}} \). This simplifies to \( v = \sqrt{\omega^2 \cdot L^2} = \omega L \).
04

Find the Time for Wave Travel

The time required for the wave to traverse the length \( L \) of the rope is given by \( t = \frac{L}{v} \). With wave speed \( v = \omega L \), substitute and solve: \( t = \frac{L}{\omega L} = \frac{1}{\omega} \).

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.

Angular Velocity
Angular velocity is crucial when dealing with rotating objects or systems, such as our whirled rope. It represents how fast an object rotates around a circle. In mathematical terms, it is expressed in radians per second. Here, it's denoted by \( \omega \).
  • The concept can be visualized as the angle an object rotates through per unit time.
  • When the rope is rotated with angular velocity \( \omega \), it influences the tension created within the rope.
  • This rotation speed plays a direct role in determining the wave speed along the rope.
Understanding the relationship between angular velocity and other physical parameters, like tension and wave speed, is essential to analyzing wave behavior.
Tension in Strings
Tension in the string arises in this context due to the centripetal force needed to keep the rope moving in a circle. The tension \( T \) is a force experienced along the length of the rope.
  • The tension depends on the mass of the rope section being considered, the angular velocity \( \omega \), and the distance from the rotation axis \( r \).
  • For our scenario, the whole mass of the rope affects the tension, resulting in the formula \( T = m \cdot \omega^2 \cdot L \).
  • Tension impacts not only the structure and stability of the rope but also how quickly waves can travel through it.
By understanding tension, we can predict how changes in the rope's speed or mass distribution affect wave progression.
Wave Speed
Wave speed is how fast a disturbance or wave travels through a medium, like a rope. For wave propagation, this speed is influenced by the tension and mass properties of the medium.
  • The foundational formula is \( v = \sqrt{\frac{T}{\mu}} \), linking tension \( T \) and linear mass density \( \mu \).
  • In this rope, substituting the determined tension and mass density gives us \( v = \omega L \), indicating that wave speed depends directly on the rope's angular velocity and length.
  • This means, for a fixed length, increasing \( \omega \) increases the wave speed, showing the dynamic nature of wave propagation in rotating strings.
Identifying wave speed helps us calculate travel times for waves and understand how they're affected by physical modifications to the system.
Linear Mass Density
Linear mass density \( \mu \) helps us understand the rope's mass distribution along its length. It's a measure of how much mass is contained in a unit length of the string, expressed as \( \mu = \frac{m}{L} \).
  • This concept is crucial when calculating wave speed because it balances the tension influences.
  • In our example, knowing \( \mu \) allows us to simplify the wave speed calculation by effectively accounting for how mass is spread over the rope's full length.
  • Higher linear mass density generally results in slower waves since there's "more mass" to move.
Understanding \( \mu \) offers insight into the properties of the medium through which waves move, affecting speed and energy transmission.

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

The B-string of a guitar is made of steel (density \(\left.7800 \mathrm{~kg} / \mathrm{m}^{3}\right),\) is \(63.5 \mathrm{~cm}\) long, and has diameter \(0.406 \mathrm{~mm}\) The fundamental frequency is \(f=247.0 \mathrm{~Hz}\) (a) Find the string tension. (b) If the tension \(F\) is changed by a small amount \(\Delta F\), the frequency \(f\) changes by a small amount \(\Delta f\). Show that $$ \frac{\Delta f}{f}=\frac{1}{2} \frac{\Delta f}{f} $$ (c) The string is tuned as in part (a) when its temperature is \(18.5^{\circ} \mathrm{C}\). Strenuous playing can make the temperature of the string rise, changing its vibration frequency. Find \(\Delta f\) if the temperature of the string rises to \(29.5^{\circ} \mathrm{C}\). The steel string has a Young's modulus of \(2.00 \times 10^{11} \mathrm{~Pa}\) and a coefficient of linear expansion of \(1.20 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1} .\) Assume that the temperature of the body of the guitar remains constant. Will the vibration frequency rise or fall?

Tuning an Instrument. A musician tunes the C.string of her instrument to a fundamental frequency of 65.4 \(\mathrm{Hz}\) . The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g}\) . (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 \(\mathrm{Hz}\) to 73.4 \(\mathrm{Hz}\) , corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

A 1.50 -m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 48.0 \(\mathrm{m} / \mathrm{s}\) . What are the wavelength and frequency of (a) the fundamental; (b) the second overtone; (c) the fourth harmonic?

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120 \(\mathrm{Hz}\) . The other end passes over a pulley and supports a \(1.50-\mathrm{kg}\) mass. The linear mass density of the rope is 0.0550 \(\mathrm{kg} / \mathrm{m}\) . (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 \(\mathrm{kg}\) ?

A \(1.50-\mathrm{m}\) string of weight 1.25 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . When you pluck the string slightly, the waves traveling up the string obey the equation $$ y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-2730 \mathrm{s}^{-1} t\right) $$ (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?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Mechanics and Materials

Read Explanation

Oscillations

Read Explanation

Circular Motion and Gravitation

Read Explanation

Atoms and Radioactivity

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.