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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.

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Most popular questions from this chapter

Imagine you are an astronaut who has landed on another planet and wants to determine the free-fall acceleration on that planet. In one of the experiments you decide to conduct, you use a pendulum \(0.50 \mathrm{~m}\) long and find that the period of oscillation for this pendulum is \(1.50 \mathrm{~s}\). What is the acceleration due to gravity on that planet?

{~A} 100 \cdot \mathrm{g}\( block hangs from a spring with \)k=5.00 \mathrm{~N} / \mathrm{m}\( At \)t=0 \mathrm{~s},\( the block is \)20.0 \mathrm{~cm}\( below the equilibrium posi. tion and moving upward with a speed of \)200, \mathrm{~cm} / \mathrm{s}\(. What is the block's speed when the displacement from equilibrium is \)30.0 \mathrm{~cm} ?$

Identical blocks oscillate on the end of a vertical spring one on Earth and one on the Moon. Where is the period of the oscillations greater? a) on Earth d) cannot be determined b) on the Moon from the information given c) same on both Farth and Moon

An object in simple harmonic motion is isochronous, meaning that the period of its oscillations is independent of their amplitude. (Contrary to a common assertion, the operation of a pendulum clock is not based on this principle. A pendulum clock operates at fixed, finite amplitude. The gearing of the clock compensates for the anharmonicity of the pendulum.) Consider an oscillator of mass \(m\) in one-dimensional motion, with a restoring force \(F(x)=-c x^{3}\) where \(x\) is the displacement from equilibrium and \(c\) is a constant with appropriate units. The motion of this ascillator is periodic but not isochronous. a) Write an expression for the period of the undamped oscillations of this oscillator. If your expression involves an integral, it should be a definite integral. You do not need to evaluate the expression. b) Using the expression of part (a), determine the dependence of the period of oscillation on the amplitude. c) Generalize the results of parts (a) and (b) to an oscillator of mass \(m\) in one-dimensional motion with a restoring force corresponding to the potential energy \(U(x)=\gamma|x| / \alpha\), where \(\alpha\) is any positive value and \(\gamma\) is a constant

A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=\) \(1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equi. librium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0 .\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0\), the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.
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