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(a) At what displacement of a SHO is the energy half kinetic and half potential? (b) What fraction of the total energy of a SHO is kinetic and what fraction potential when the displacement is one third the amplitude?

Short Answer

Expert verified
(a) Displacement is \(\frac{A}{\sqrt{2}}\). (b) Fractions are \(\frac{4}{9}\) kinetic and \(\frac{1}{9}\) potential.

Step by step solution

01

Understanding Total Energy of SHO

In simple harmonic motion (SHO), the total mechanical energy, \(E\), is conserved and is the sum of kinetic energy, \(K\), and potential energy, \(U\). It's given by:\[ E = K + U = \frac{1}{2} m v^2 + \frac{1}{2} k x^2 \]where \(m\) is the mass of the object, \(v\) is the velocity, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. The maximum potential energy is \(\frac{1}{2} k A^2\), where \(A\) is the amplitude.
02

Setting Up Equation for Half Kinetic and Half Potential Energy

We want to find the displacement \(x\) for which the kinetic energy equals potential energy, meaning:\[ K = U = \frac{1}{2}E \]Therefore, \(\frac{1}{2} m v^2 = \frac{1}{2} k x^2\). Given that total energy \(E = \frac{1}{2} k A^2\), we can substitute and solve for \(x\):\[ \frac{1}{2} k x^2 = \frac{1}{2} \cdot \frac{1}{2} k A^2 \] \[ x^2 = \frac{1}{2} A^2 \]\[ x = \frac{A}{\sqrt{2}} \]
03

Evaluating Energy Fractions at Displacement of One Third of Amplitude

Next, calculate the energies when the displacement \(x = \frac{A}{3}\). Plug \(x\) into the potential energy equation:\[ U = \frac{1}{2} k \left(\frac{A}{3}\right)^2 = \frac{1}{2} k \frac{A^2}{9} = \frac{1}{18} k A^2 \]The total energy \(E = \frac{1}{2} k A^2\), hence the kinetic energy \(K\) is:\[ K = E - U = \frac{1}{2} k A^2 - \frac{1}{18} k A^2 = \frac{8}{18} k A^2 = \frac{4}{9} k A^2 \]Therefore, the fraction of kinetic energy is \(\frac{4}{9}\) and the fraction of potential energy is \(\frac{1}{9}\).
04

Summary of Results

For part (a), the displacement at which energy is half kinetic and half potential is \(\frac{A}{\sqrt{2}}\). For part (b), when the displacement is \(\frac{A}{3}\), the fractions of total energy are \(\frac{4}{9}\) kinetic and \(\frac{1}{9}\) potential.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy in simple harmonic motion (SHO) can be a bit like watching a swing in action. Essentially, when an object in SHO is at its maximum speed, all of its energy is kinetic. This is because it's moving, and kinetic energy is due to motion. The mathematical expression for kinetic energy, \[ K = \frac{1}{2} m v^2 \] makes it clear that it depends on both the mass (\( m \)) and the velocity (\( v \)) of the object. As the object moves back and forth in its oscillation, its kinetic energy changes.
  • At the equilibrium position, kinetic energy is maximum because velocity is at its peak.
  • At maximum displacement (where the object stops momentarily), kinetic energy is zero as there is no movement.
The key takeaway is that the kinetic energy of a system in SHO varies with its speed and position. No need to panic though—calculating this energy simply involves using the object's speed and mass. In exercises like the one discussed, kinetic energy becomes crucial as it transforms interchangeably into potential energy.
Potential Energy
When discussing potential energy in the context of simple harmonic motion (SHO), think of it as the stored energy that the system possesses due to its position. It's like pulling back on a spring, knowing it will leap forward once you let go. The potential energy in SHO is expressed as\[ U = \frac{1}{2} k x^2 \]where \(k\) represents the spring constant and \(x\) denotes the displacement from the equilibrium position.
  • At maximum displacement, potential energy is at its highest because the system is most stretched or compressed.
  • In contrast, at the equilibrium point (or mean position), potential energy is minimal because the spring is neither stretched nor compressed.
To solve related exercises, you often need to understand how potential and kinetic energy exchange places as the object oscillates. For example, when potential energy is at its half value of the total energy, you can directly calculate the accompanying displacement, as in the exercise: \( x = \frac{A}{\sqrt{2}} \). Recognizing these transitions can help students not only solve problems but also intuitively understand how potential and kinetic energies complement each other.
Mechanical Energy Conservation
The principle of mechanical energy conservation in a simple harmonic oscillator is like a dance of different energies, where the total energy remains constant throughout the motion. This means every bit of potential energy that decreases is converted back into kinetic energy, and vice versa. The relationship is simple:\[ E = K + U \]Here, \( E \) represents the total mechanical energy, which doesn't change, even as \( K \) and \( U \) shuffle values back and forth.
  • This conservation means if you know the total energy, analyzing either kinetic or potential energy at any position helps determine the other.
  • In exercises, this concept helps find unknowns like displacement or energy fractions by setting the known total energy equal to the sum of \( K \) and \( U \).
For example, finding when the energy splits evenly between kinetic and potential is a practical demonstration of this principle. In SHO, the equal mix happens at the displacement \( x = \frac{A}{\sqrt{2}} \), as demonstrated in the step-by-step solution. By grasping these interactions, students can better handle problems involving energy conservation, helping them connect dots between practical motion and theoretical principles.

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Most popular questions from this chapter

(II) A physical pendulum consists of an 85 -cm-long, 240 -g-mass, uniform wooden rod hung from a nail near one end (Fig, 38 ). The motion is damped because of friction in the pivot; the damping force is approximately proportional to \(d \theta / d t .\) The rod is set in oscillation by displacing it \(15^{\circ}\) from its equilibrium position and releasing it. After 8.0 \(\mathrm{s}\) the amplitude of the oscillation has been reduced to \(5.5^{\circ} .\) If the angular displacement can be written as \(\theta=A e^{-\gamma t} \cos \omega^{\prime} t,\) find \((a) \gamma,(b)\) the approximate period of the motion, and \((c)\) how long it takes for the amplitude to be reduced to \(\frac{1}{2}\) of its original value.

A bungee jumper with mass \(65.0 \mathrm{~kg}\) jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in \(43.0 \mathrm{~s}\). He finally comes to rest \(25.0 \mathrm{~m}\) below the level of the bridge. Estimate the spring stiffness constant and the unstretched length of the bungee cord assuming SHM.

A vertical spring of spring constant \(115 \mathrm{~N} / \mathrm{m}\) supports a mass of \(75 \mathrm{~g}\). The mass oscillates in a tube of liquid. If the mass is initially given an amplitude of \(5.0 \mathrm{~cm},\) the mass is observed to have an amplitude of \(2.0 \mathrm{~cm}\) after \(3.5 \mathrm{~s}\). Estimate the damping constant \(b\). Neglect buoyant forces.

(II) A mass \(m\) on a frictionless surface is attached to a spring with spring constant \(k\) as shown in Fig. \(47 .\) This mass-spring system is then observed to execute simple harmonic motion with a period \(T .\) The mass \(m\) is changed several times and the associated period \(T\) is measured in each case, generating the following data Table: (a) Starting with \(\mathrm{Eq} .7 \mathrm{b}\) , show why a graph of \(T^{2}\) vs. \(m\) is expected to yield a straight line. How can \(k\) be determined from the straight line's slope? What is the line's \(y\) -intercept expected to be? (b) Using the data in the Table, plot \(T^{2}\) vs. \(T^{2}\) vs. \(T^{2}\) vs. and show that this graph yields a straight line. Determine the slope and (nonzero) y-intercept. (c) Show that a nonzero \(y\) -intercept can be expected in our plot theoretically if, rather than simply using \(m\) for the mass in Eq. \(7 b,\) we use \(m+m_{0},\) where \(m_{0}\) is a constant. That is, repeat part \((a)\) using \(m+m_{0}\) for the mass in Eq. 7 \(\mathrm{b}\) . Then use the result of this analysis to determine \(k\) and \(m_{0}\) from your graph's slope and \(y\) -intercept. (d) Offer a physical interpretation for \(m_{0},\) a mass that appears to be oscillating in addition to the attached mass \(m .\) \(T=2 \pi \sqrt{\frac{m}{k}}\)

(II) An object of unknown mass \(m\) is hung from a vertical spring of unknown spring constant \(k,\) and the object is observed to be at rest when the spring has extended by 14 \(\mathrm{cm} .\) The object is then given a slight push and executes SHM. Determine the period \(T\) of this oscillation.

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