/*! 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 The upper end of a 3.80-m-long s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The upper end of a 3.80-m-long steel wire is fastened to the ceiling, and a 54.0-kg object is suspended from the lower end of the wire. You observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. What is the mass of the wire?

Short Answer

Expert verified
The mass of the wire is approximately 0.34 kg.

Step by step solution

01

Calculate the Speed of the Pulse

The formula for speed is \( v = \frac{d}{t} \), where \( d \) is the distance and \( t \) is the time. Here, \( d = 3.80 \) m and \( t = 0.0492 \) s.\[ v = \frac{3.80}{0.0492} \approx 77.24 \text{ m/s} \]
02

Relate Pulse Speed, Tension, and Mass Density

The speed \( v \) of a wave on a string is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the wire and \( \mu \) is the mass per unit length (linear density). We rearrange this to find \( \mu \):\[ \mu = \frac{T}{v^2} \]
03

Calculate the Tension in the Wire

The tension \( T \) in the wire is equal to the weight of the object suspended, which can be calculated as \( T = mg \), where \( m \) is mass and \( g \) is acceleration due to gravity \( 9.81 \text{ m/s}^2 \).\[ T = 54.0 \times 9.81 = 529.74 \text{ N} \]
04

Solve for Mass Density

Substituting the known values into the equation for \( \mu \):\[ \mu = \frac{529.74}{77.24^2} \approx 0.0889 \text{ kg/m} \]
05

Calculate the Mass of the Wire

The mass of the wire \( M \) can be found using the linear density \( \mu \) and the total length of the wire. The formula is \( M = \mu \times L \), where \( L \) is the length of the wire (3.80 m).\[ M = 0.0889 \times 3.80 = 0.33782 \text{ kg} \]

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 91Ó°ÊÓ!

Key Concepts

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

Transverse Waves
Transverse waves are waves where the displacement of the medium is perpendicular to the direction of wave propagation. This concept is integral to understanding wave mechanics because it explains how energy travels through different media.
  • Examples of transverse waves include light waves or vibrations in a string.
  • They are characterized by their crests (the highest points) and troughs (the lowest points).
Transverse waves in a wire, as in the exercise, allow us to study the motion and properties of waves such as speed, frequency, and amplitude. For the wire in question, a transverse pulse is created, and its travel time over a known distance helps us compute the wave's speed. This wave speed is essential in determining other factors like tension and linear density in the wire.
Understanding transverse waves enables you to comprehend how mechanical waves can convey energy without transporting matter. It also provides insights into the elastic properties of the medium in which they travel. This knowledge is crucial for analyzing various physical systems and engineering structures.
Linear Density
Linear density, denoted as \( \mu \), is a measure of mass per unit length. It's crucial in determining how a wave travels through a medium such as a wire.
  • The formula \( \mu = \frac{m}{L} \) describes linear density, where \( m \) is mass and \( L \) is length.
  • It's a key factor in calculating the speed of waves in a medium, especially when related to tension.
In the given problem, calculating linear density aids in understanding how the physical properties of the wire affect the behavior of the wave. The exercise demonstrates how linear density can be calculated using the tension and speed of the wave in the wire. By knowing the linear density, we can ascertain the mass of the wire based on its length and other dynamic properties. Comprehending linear density and its effects helps in designing and analyzing systems like musical instruments and suspension bridges, where wave mechanics play a vital role.
Tension in a Wire
The concept of tension in a wire refers to the force transmitted through a string, wire, or rope when it is pulled tight by forces acting from opposite ends.
  • In physics, tension is typically measured in newtons (N).
  • In our exercise, tension is determined by the weight of an object attached to the wire:
  • \( T = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity.
Tension is a critical element in determining the wave speed in a medium. For the given problem, it's essential to calculate the tension accurately, as it directly influences the wave's speed through the equation \( v = \sqrt{\frac{T}{\mu}} \). By knowing the tension and the wave speed, one can calculate the linear density, which is then used to ascertain the mass of the entire wire.
Tension is a widely applicable concept in various scientific and engineering disciplines, including material science and mechanical engineering. Understanding how tension affects wave motion can inform the design and analysis of structures or systems that require precision and reliability.

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

By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is 0.026 W/m\(^2\) at a distance of 4.3 m from the source. (a) What is the intensity at a distance of 3.1 m from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

The speed of sound in air at 20\(^\circ\)C is 344 m/s. (a) What is the wavelength of a sound wave with a frequency of 784 Hz, corresponding to the note G\(_5\) on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher (twice the frequency) than the note in part (a)?

Transverse waves on a string have wave speed 8.00 m/s, amplitude 0.0700 m, and wavelength 0.320 m. The waves travel in the \(-x\)-direction, and at \(t = 0\) the \(x = 0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at \(x = 0.360\) m at time \(t = 0.150\) s. (d) How much time must elapse from the instant in part (c) until the particle at \(x = 0.360\) m next has maximum upward displacement?

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the \(+x\)-axis and is fixed at \(x = 0\). (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?

For a string stretched between two supports, two successive standing-wave frequencies are 525 Hz and 630 Hz. There are other standing-wave frequencies lower than 525 Hz and higher than 630 Hz. If the speed of transverse waves on the string is 384 m/s, what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.

See all solutions

Recommended explanations on Physics Textbooks

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.