Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 57

A mass \(m\) is attached to a spring with a spring constant of \(k\) and set into simple harmonic motion. When the mass has half of its maximum kinetic energy, how far away from its equilibrium position is it, expressed as a fraction of its maximum displacement?

Short Answer

Expert verified
Answer: The distance from the equilibrium position when the mass has half of its maximum kinetic energy is approximately 0.707 times the maximum displacement.

Step by step solution

01

Write the equation for total energy in simple harmonic motion

For an object in simple harmonic motion, the total energy (E) is the sum of its kinetic energy (K) and potential energy (U). The total energy can be expressed as: E = K + U
02

Write down the expressions for kinetic and potential energy

Kinetic energy K can be expressed as: K = \frac{1}{2}mv^2 where v is the velocity of the mass. Potential energy U can be expressed as: U = \frac{1}{2}kx^2 where x is the displacement from the equilibrium position.
03

Introduce the given constraint of half-maximum kinetic energy

We are given that K is half of its maximum value. The maximum kinetic energy can be expressed as: K_max = \frac{1}{2}kA^2 where A is the maximum displacement (amplitude). Since K is half of K_max: K = \frac{1}{2}K_max = \frac{1}{4}kA^2
04

Substitute the expressions for K and U in the equation for E

We have E as K + U, with K and U defined in steps 2 and 3. Substituting the expressions for K and U, we get: E = \frac{1}{4}kA^2 + \frac{1}{2}kx^2
05

Express the maximum displacement A in terms of total energy E

Applied force at maximum displacement is given by Hooke's law: F = -kA Energy at maximum displacement is given by potential energy U, since there is no kinetic energy at that point: E = \frac{1}{2}kA^2 Solving for A, we get: A^2 = \frac{2E}{k}
06

Substitute A^2 back into the equation from step 4, and solve for x

Replacing A^2 in the equation from step 4, we get: E = \frac{1}{4}k (\frac{2E}{k}) + \frac{1}{2}kx^2 => E = \frac{1}{2}E + \frac{1}{2}kx^2 Now, solving for x^2: x^2 = \frac{1}{k} (E - \frac{1}{2}E) = \frac{1}{2} \frac{2E}{k} = \frac{1}{2}A^2
07

Express the displacement as a fraction of maximum displacement and find its square root

Now that we have x^2 in terms of A^2, we can express x as a fraction of A: \frac{x}{A} = \sqrt{\frac{x^2}{A^2}} = \sqrt{\frac{1}{2}} So, when the mass has half of its maximum kinetic energy, it is: x = \sqrt{\frac{1}{2}}A away from its equilibrium position expressed as a fraction of its maximum displacement A. This means that the mass is approximately 0.707 times the maximum displacement away from the equilibrium position when it has half of its maximum kinetic energy.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Pendulum A has a bob of mass \(m\) hung from a string of length \(I_{i}\) pendulum \(B\) is identical to \(A\) except its bob has mass \(2 m\). Compare the frequencies of small oscillations of the two pendulums.

A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass \(M\) and length \(L\) that is pivoted freely about one end. with a solid sphere of the same mass, \(M,\) and a radius of \(L / 2\) centered about the free end of the rod. a) Obtain an expression for the moment of inertia of the pendulum about its pivot point as a function of \(M\) and \(L\). b) Obtain an expression for the period of the pendulum for small oscillations.

A block of wood of mass \(55.0 \mathrm{~g}\) floats in a swimming pool, oscillating up and down in simple harmonic motion with a frequency of \(3.00 \mathrm{~Hz}\). a) What is the value of the effective spring constant of the water? b) A partially filled water bottle of almost the same size and shape as the block of wood but with mass \(250 . g\) is placed on the water's surface. At what frequency will the bottle bob up and down?

A horizontal tree branch is directly above another horizontal tree branch. The elevation of the higher branch is \(9.65 \mathrm{~m}\) above the ground, and the elevation of the lower branch is \(5.99 \mathrm{~m}\) above the ground. Some children decide to use the two branches to hold a tire swing. One end of the tire swing's rope is tied to the higher tree branch so that the bottom of the tire swing is \(0.47 \mathrm{~m}\) above the ground. This swing is thus a restricted pendulum. Start. ing with the complete length of the rope at an initial angle of \(14.2^{\circ}\) with respect to the vertical, how long does it take a child of mass \(29.9 \mathrm{~kg}\) to complete one swing back and forth?

A cylindrical can of diameter \(10.0 \mathrm{~cm}\) contains some ballast so that it floats vertically in water. The mass of can and ballast is \(800.0 \mathrm{~g}\), and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\) The can is lifted \(1.00 \mathrm{~cm}\) from its equilibrium position and released at \(t=0 .\) Find its vertical displacement from equilibrium as a function of time. Determine the period of the motion. Ignore the damping effect due to the viscosity of the water.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Energy Physics

Read Explanation

Scientific Method Physics

Read Explanation

Rotational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Radiation

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.

Sign-up for free