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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

An \(80.0-\mathrm{kg}\) bungee jumper is enjoying an afternoon of jumps. The jumper's first oscillation has an amplitude of \(10.0 \mathrm{~m}\) and a period of \(5.00 \mathrm{~s}\). Treating the bungee cord as spring with no damping, calculate each of the following: a) the spring constant of the bungee cord. b) the bungee jumper's maximum speed during the ascillation, and c) the time for the amplitude to decrease to \(2.00 \mathrm{~m}\) (with air resistance providing the damping of the oscillations at \(7.50 \mathrm{~kg} / \mathrm{s})\)

With the right choice of parameters, a damped and driven physical pendulum can show chaotic motion, which is sensitively dependent on the initial conditions. Which statement about such a pendulum is true? a) Its long-term behavior can be predicted. b) Its long-term behavior is not predictable. c) Its long-term behavior is like that of a simple pendulum of equivalent length. d) Its long-term behavior is like that of a conical pendulum. e) None of the above is true.

A mass of \(10.0 \mathrm{~kg}\) is hanging by a steel wire \(1.00 \mathrm{~m}\) long and \(1.00 \mathrm{~mm}\) in diameter. If the mass is pulled down slightly and released, what will be the frequency of the resulting oscillations? Young's modulus for steel is \(2.0 \cdot 10^{11} \mathrm{~N} / \mathrm{m}^{2}\)

A mass, \(M=1.6 \mathrm{~kg}\), is attached to a wall by a spring with \(k=578 \mathrm{~N} / \mathrm{m}\). The mass slides on a frictionless floor. The spring and mass are immersed in a fluid with a damping constant of \(6.4 \mathrm{~kg} / \mathrm{s}\). A horizontal force, \(\mathrm{F}(\mathrm{t})=\mathrm{F}_{\mathrm{d}} \cos \left(\omega_{\mathrm{d}} \mathrm{t}\right)\) where \(F_{d}=52 \mathrm{~N},\) is applied to the mass through a knob, caus. ing the mass to oscillate back and forth. Neglect the mass of the spring and of the knob and rod. At approximately what frequency will the amplitude of the mass's oscillation be greatest, and what is the maximum amplitude? If the driving frequency is reduced slightly (but the driving amplitude remains the same), at what frecuency will the amplitude of the mass's ascillation be half of the maximum amplitude?

A spring is hanging from the ceiling with a mass attached to it. The mass is pulled downward, causing it to oscillate vertically with simple harmonic motion. Which of the following will increase the frequency of oscillation? a) adding a second, identical spring with one end attached to the mass and the other to the ceiling b) adding a second, identical spring with one end attached to the mass and the other to the floor c) increasing the mass d) adding both springs, as described in (a) and (b)
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