/*! 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 67 The formula for the speed of tra... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 67

The formula for the speed of transverse waves on a spring is the same as for a string. (a) A spring is stretched to a length much greater than its relaxed length. Explain why the tension in the spring is approximately proportional to the length. (b) A wave takes 4.00 s to travel from one end of such a spring to the other. Then the length is increased \(10.0 \% .\) Now how long does a wave take to travel the length of the spring? [Hint: Is the mass per unit length constant?]

Short Answer

Expert verified
Answer: The tension in a spring is approximately proportional to its length when stretched because of Hooke's Law, which states that the spring force is proportional to the displacement from the equilibrium position. After the length of the spring is increased by 10%, the wave takes 4.4 seconds to travel from one end to another.

Step by step solution

01

Understand Hooke's Law and spring force

According to Hooke's Law, the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position: \(F = -kx\), where \(F\) is the spring force, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. When a spring is stretched to a length much greater than its relaxed length, the tension in the spring becomes approximately equal to the spring force (\(T \approx F\)).
02

Apply Hooke's Law to spring tension

Since the spring force is proportional to the displacement from the equilibrium position (\(F \propto x\)), we can also say that the tension in the spring is approximately proportional to the displacement (\(T \approx F \propto x\)). Thus, when the spring is stretched to a length much greater than its relaxed length, the tension in the spring is approximately proportional to the length. #b) A wave takes 4.00 s to travel from one end of such a spring to the other. Then the length is increased \(10.0 \% .\) Now how long does a wave take to travel the length of the spring?#
03

Use given information

From the problem, we know that the wave on a spring takes \(4.00\,\text{s}\) to travel from one end to the other. We are also given that the length is increased by \(10.0 \%\) or \(1.10 L\), where \(L\) is the initial length of the spring.
04

Apply wave speed equation

The wave speed equation is given by \(v = \frac{d}{t}\), where \(v\) is the wave speed, \(d\) is the distance traveled, and \(t\) is the time taken. We can write this equation for the initial length (\(L\)) and the increased length (\(1.10 L\)) to find the new time taken for a wave to travel the length of the spring. Let \(v_1 = \frac{L}{t_1}\) and \(v_2 = \frac{1.10L}{t_2}\), where \(t_1 = 4.00\,\text{s}\) and \(t_2\) is the time taken to travel the increased length. Since the mass per unit length is constant, the wave speed remains the same, that is, \(v_1 = v_2\).
05

Solve for the new time taken

Using the wave speed equation, we have: \(\frac{L}{t_1} = \frac{1.10L}{t_2}\) Now we solve for \(t_2\): \(t_2 = t_1 \cdot \frac{1.10L}{L}\) Substitute the given values: \(t_2 = 4.00\,\text{s} \cdot \frac{1.10L}{L}\) Simplify the expression: \(t_2 = 4.4\,\text{s}\) So, after the length of the spring is increased by \(10.0 \%\), the wave takes \(4.4\,\text{s}\) to travel from one end to the other.

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

(a) Use a graphing calculator or computer graphing program to plot \(y\) versus \(x\) for the function $$ y(x, t)=(5.0 \mathrm{cm})[\sin (k x-\omega t)+\sin (k x+\omega t)] $$ for the times \(t=0,1.0 \mathrm{s},\) and \(2.0 \mathrm{s}\). Use the values \(k=\pi /(5.0 \mathrm{cm})\) and \(\omega=(\pi / 6.0) \mathrm{rad} / \mathrm{s} .\) (b) Is this a traveling wave? If not, what kind of wave is it?
Light visible to humans consists of electromagnetic waves with wavelengths (in air) in the range \(400-700 \mathrm{nm}\) $\left(4.0 \times 10^{-7} \mathrm{m} \text { to } 7.0 \times 10^{-7} \mathrm{m}\right) .$ The speed of light in air is \(3.0 \times 10^{8} \mathrm{m} / \mathrm{s} .\) What are the frequencies of electromagnetic waves that are visible?
A sound wave with intensity \(25 \mathrm{mW} / \mathrm{m}^{2}\) interferes constructively with a sound wave that has an intensity of $15 \mathrm{mW} / \mathrm{m}^{2} .$ What is the intensity of the superposition of the two? (tutorial: superposition)

The Sun emits electromagnetic waves (including light) equally in all directions. The intensity of the waves at Earth's upper atmosphere is \(1.4 \mathrm{kW} / \mathrm{m}^{2} .\) At what rate does the Sun emit electromagnetic waves? (In other words, what is the power output?)
The longest "string" (a thick metal wire) on a particular piano is $2.0 \mathrm{m}\( long and has a tension of \)300.0 \mathrm{N} .$ It vibrates with a fundamental frequency of \(27.5 \mathrm{Hz}\). What is the total mass of the wire?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Scientific Method Physics

Read Explanation

Turning Points in Physics

Read Explanation

Wave Optics

Read Explanation

Physics of Motion

Read Explanation

Physical Quantities And Units

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.