/*! 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 17 A thin, 75.0 -cm wire has a mass... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 17

A thin, 75.0 -cm wire has a mass of 16.5 g. One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 \(\mathrm{cm}\) makes 875 vibrations per second? (b) How fast would this wave travel?

Short Answer

Expert verified
(a) 18.72 N, (b) 29.14 m/s

Step by step solution

01

Convert Units

First, convert the given measurements to standard SI units. The length of the wire is 75.0 cm, which is 0.75 meters. The mass of the wire is 16.5 g, which is 0.0165 kg. The wavelength is 3.33 cm, which is 0.0333 meters.
02

Calculate Linear Mass Density

The linear mass density (mass per unit length) of the wire is calculated using the formula \( \mu = \frac{m}{L} \), where \( m \) is the mass and \( L \) is the length of the wire. Thus, \( \mu = \frac{0.0165}{0.75} = 0.022 \, \text{kg/m} \).
03

Find Wave Speed

Use the frequency and wavelength to find the wave speed \( v \) using the relation \( v = f \lambda \), where \( f \) is the frequency and \( \lambda \) is the wavelength. The frequency is 875 Hz. Thus, the wave speed is \( v = 875 \times 0.0333 = 29.1375 \text{ m/s} \).
04

Calculate Tension

The tension \( T \) in the wire can be found using the formula \( v = \sqrt{\frac{T}{\mu}} \). Rearrange to solve for \( T \): \( T = \mu v^2 \). Substituting the known values gives \( T = 0.022 \times (29.1375)^2 \), resulting in a tension of approximately 18.72 Newtons.

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 how to calculate the speed of a wave is crucial in physics, especially for waves traveling on a medium such as a wire. Wave speed can be determined using the relationship between wave frequency and wavelength.
  • Frequency is the number of complete wave cycles that pass a point in one second; it is measured in Hertz (Hz).
  • Wavelength is the distance between successive crests of a wave, measured in meters (m).
To find wave speed, use the formula: \[ v = f \lambda \]where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. Imagine a wave with a frequency of 875 vibrations per second and a wavelength of 0.0333 meters. By plugging these values into the formula, the wave speed can be calculated:\[ v = 875 \, \times \, 0.0333 \, = \, 29.1375 \, \text{m/s} \]. This equation is fundamental because it connects the physical properties of the wave (wavelength and frequency) to how fast the wave travels.
Linear Mass Density
Linear mass density plays a significant role in understanding transverse waves on a wire. It refers to the mass of the wire per unit length.
  • Expressed mathematically as \( \mu = \frac{m}{L} \), where \( m \) is the mass of the wire and \( L \) is its length.
  • Measured in kilograms per meter (kg/m).
In our example, the wire has a length of 0.75 meters and a mass of 0.0165 kilograms. By substituting these values into the formula, we can determine the linear mass density:\[ \mu = \frac{0.0165}{0.75} \, = \, 0.022 \, \text{kg/m} \]. This value is crucial because it affects the wave speed and tension in the wire. Knowing the linear mass density helps us analyze how forces act on the wire during wave propagation.
Tension in Wire
Tension is a force that is applied along the length of the wire, and it affects how waves travel through the medium. Calculating the tension is necessary for understanding wave dynamics.
  • The tension \( T \) affects the speed \( v \) of the wave and depends on the linear mass density \( \mu \).
  • To find tension, use the formula \( v = \sqrt{\frac{T}{\mu}} \), which can be rearranged to solve for \( T \).
Rewriting the formula to find tension gives us:\[ T = \mu v^2 \]. Inserting the known values \( \mu = 0.022 \, \text{kg/m} \) and \( v = 29.1375 \, \text{m/s} \), we find:\[ T = 0.022 \, \times \, (29.1375)^2 \, \approx \, 18.72 \, \text{N} \].This calculation illustrates how changes in tension influence the wave speed. Understanding this relationship helps in designing systems with controlled wave properties.

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 portion of the string of a certain musical instrument between the bridge and upper end of the finger board (that part of the string that is free to vibrate) is 60.0 \(\mathrm{cm}\) long. and this length of the string has mass 2.00 g. The string sounds an \(\mathrm{A}_{4}\) note \((440 \mathrm{Hz})\) when played. (a) Where must the player put a finger (what distance \(x\) from bridge) to play \(\mathbf{a}\) Ds note \((587 \quad \mathrm{Hz}) ? \quad\) (See Fig. 15.36 . For both the A_{4} and \(D_{5}\) notes, the string vibrates in its fundamental mode. (b) Without retuning, is it possible to play a G \(_{4}\) note \((392 \mathrm{Hz})\) on this string? Why or why not?

Adjacent antinodes of a standing wave on a string are 15.0 \(\mathrm{cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 \(\mathrm{cm}\) and period 0.0750 \(\mathrm{s}\) . The string lies along the \(+x\) -axis and is fixed at \(x=0 .(\text { a) How far }\) apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

A piano wire with mass 3.00 \(\mathrm{g}\) and length 80.0 \(\mathrm{cm}\) is stretched with a tension of 25.0 \(\mathrm{N}\) . A wave with frequency 120.0 \(\mathrm{Hz}\) and amplitude 1.6 \(\mathrm{mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

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

Wave Equation and Standing Waves. (a) Prove by direct substitution that \(y(x, t)=\left(A_{\text { sw }} \sin k x\right)\) sin \(\omega t\) is a solution of the wave equation, Eq. \((15.12),\) for \(v=\omega / k .\) (b) Explain why the relationship \(v=\omega / k\) for traveling waves also applies to standing waves.
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Fundamentals of Physics

Read Explanation

Electrostatics

Read Explanation

Quantum Physics

Read Explanation

Oscillations

Read Explanation

Geometrical and Physical Optics

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.