/*! 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 27 A transverse wave with 3.0 -cm a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 27

A transverse wave with 3.0 -cm amplitude and 75 -cm wavelength propagates at \(6.7 \mathrm{m} / \mathrm{s}\) on a stretched spring with mass per unit length \(170 \mathrm{g} / \mathrm{m} .\) Find the spring tension.

Short Answer

Expert verified
After solving, we get the spring tension 'T' to be approximately \(7.59 \mathrm{N}\).

Step by step solution

01

Calculating the wave speed v

Firstly, it's important to convert all units to standard ones. Therefore, we convert the wave speed to meters per second. In this case, we already have the wave speed, given to be \(6.7 \mathrm{m/s}\), so no calculations are necessary.
02

Calculating the mass per unit length \(\mu\)

Next, we need to convert the mass per unit length to the standard units. Since it's given as \(170 \mathrm{g/m}\), we need to convert grams into kilograms, therefore, \(\mu = 170 \mathrm{g/m} = 0.17 \mathrm{kg/m}\).
03

Use the wave speed equation to find the tension

Substitute our known values of 'v' and '\mu' back into the wave speed equation, and solve for 'T'. We can rearrange the equation to make 'T' the subject and then substitute our values: \(T = v^{2} * \mu\), therefore \( T = (6.7 \mathrm{m/s})^{2} * 0.17 \mathrm{kg/m} \).

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!

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

When a wave source moves relative to the medium, a stationary observer measures changes in both wavelength and frequency. But when the observer moves and the source is stationary, only the frequency changes. Why the difference?

The main cables supporting New York's George Washington Bridge have a mass per unit length of \(4100 \mathrm{kg} / \mathrm{m}\) and are under 250-MN tension. At what speed would a transverse wave propagate on these cables?

Analysis of waves in shallow water (depth much less than wavelength) yields the following wave equation: $$\frac{\partial^{2} y}{\partial x^{2}}=\frac{1}{g h} \frac{\partial^{2} y}{\partial t^{2}}$$ where \(h\) is the water depth and \(g\) the gravitational acceleration. Give an expression for the wave speed.

Medical ultrasound uses frequencies around \(10^{7} \mathrm{Hz}\), far above the range of the human ear. In what sense are these waves "sound"?

An ideal spring is stretched to a total length \(L_{1} .\) When that length is doubled, the speed of transverse waves on the spring triples. Find an expression for the unstretched length of the spring.
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Fluids

Read Explanation

Rotational Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Space 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.