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Chapter 12: Problem 62

An object on a spring slides on a horizontal frictionless surface with simple harmonic motion. Determine where the object's kinetic energy and potential energy are the same. Assume the maximum displacement of the object is \(A\) so it oscillates between \(+A\) and \(-A\).

Short Answer

Expert verified
The position where the kinetic and potential energy of the object are the same, given maximum displacement \(A\), mass \(m\), spring constant \(k\), and angular frequency \(\omega\), is \(x = 卤A \sqrt{\frac{m\omega^2}{k}} \sin (\omega t + \phi)\).

Step by step solution

01

Set up Kinetic and Potential Energy Equations

Knowing that the energy is conserved, the sum of potential energy and kinetic energy in a simple harmonic motion is constant. This means \(\frac{1}{2} m v^2 + \frac{1}{2} k x^2 =\text{constant} = \frac{1}{2} k A^2\). This constant is chosen such that it represents the total energy when the object is at its maximum displacement (i.e., at \(卤A\), where it has maximum potential energy and no kinetic energy.)
02

Equate Kinetic and Potential Energy

To find the position(s) where the kinetic and potential energy of the object are equal, we set \(\frac{1}{2} m v^2 = \frac{1}{2} k x^2\). Simplifying, we get the relationship \(m v^2 = k x^2\).
03

Express velocity in terms of displacement

From the equation of motion for simple harmonic motion \(x=A \cos (\omega t + \phi)\), we can find velocity by differentiating displacement with respect to time: \(v = -A \omega \sin (\omega t + \phi)\). Therefore, the kinetic energy becomes \(\frac{1}{2} m A^2 \omega^2 \sin^2 (\omega t + \phi)\) where \(\omega\) is the angular frequency.
04

Substitute Velocity and Solve

Substitute the expression of \(v^2\) from step 3 into the equation \(m v^2 = k x^2\) we obtained in step 2. We have \( m A^2 \omega^2 \sin^2 (\omega t + \phi) = k x^2\). Solving for \(x\), we find \(x = 卤A \sqrt{\frac{m\omega^2}{k}} \sin (\omega t + \phi)\).

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Key Concepts

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

Kinetic Energy
In simple harmonic motion, kinetic energy is associated with the object's velocity. It is the energy due to motion and is given by the equation \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. At any point during the oscillation, the kinetic energy depends on how fast the object is moving.

One important aspect of simple harmonic motion is that kinetic energy changes as the object moves. It is zero when the object reaches the maximum displacement, either at \(+A\) or \(-A\), as the object momentarily stops before reversing its direction.

As the object moves towards the equilibrium position, its speed, and hence kinetic energy, increases. The maximum kinetic energy occurs when the object passes through the equilibrium point, where the velocity is maximal.
Potential Energy
Potential energy in simple harmonic motion is primarily due to the position of the object in relation to its equilibrium position. It is given by the equation \( PE = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

At maximum displacement, which is either at \(+A\) or \(-A\), the energy of the system is entirely potential. This is because the object has been stretched or compressed to its maximum extent and is momentarily at rest.

As the object moves toward the equilibrium point, potential energy decreases while kinetic energy increases. When the object passes through the equilibrium, potential energy is zero since \( x = 0 \), and the spring is neither compressed nor stretched.
Energy Conservation
A key principle in simple harmonic motion is the conservation of energy. Throughout the motion, the total mechanical energy remains constant, provided there are no energy losses due to factors like friction.

The total mechanical energy, which is the sum of kinetic and potential energies, is given by \( E = \frac{1}{2} k A^2 \). This illustrates that at maximum displacement, the entire energy is potential. Conversely, when the object is at the equilibrium position, the energy is purely kinetic.

The point at which kinetic and potential energies are equal occurs when the displacement \( x \) is \( \pm A/\sqrt{2} \). At these positions, the energy is equally divided between kinetic and potential energy, illustrating the elegant interplay of these energies in simple harmonic motion.

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Most popular questions from this chapter

An oscillating system has a natural frequency of \(50 \mathrm{rad} / \mathrm{s}\). The damping coefficient is \(2.0 \mathrm{~kg} / \mathrm{s}\). The system is driven by a force \(F(t)=(100 \mathrm{~N}) \cos ((50 \mathrm{rad} / \mathrm{s}) t)\). What is the amplitude of the oscillations? SSM

The potential energy of an object on a spring is \(2.4 \mathrm{~J}\) at a location where the kinetic energy is \(1.6 \mathrm{~J}\). If the amplitude of the simple harmonic motion is \(20 \mathrm{~cm}\), (a) calculate the spring constant and (b) find the largest force that it experiences.

The application of an external force on a simple pendulum can create many different outcomes, depending on how frequently the force is applied. Explain what will happen to the amplitude of the motion if an external force is applied to a simple pendulum at the same frequency as the natural frequency of the pendulum.

A \(60\)-cm-long thin rod of copper has a radius \(r\) of \(0.4 \mathrm{~cm}\) (the density of copper is \(8.92 \mathrm{~g} / \mathrm{cm}^{3}\) ). The rod is suspended from a thin wire that is welded to the exact center of the copper rod. The wire is also made of copper and has a length of \(20 \mathrm{~cm}\) and a cross-sectional diameter of \(1 \mathrm{~mm}\). The rod is displaced from the equilibrium position and the torque on the thin wire causes it to twist the rod back and forth in harmonic motion. This is a torsion pendulum (Figure 12-33). The torque acts on the rod according to following equation: $$ \tau=-K \theta $$ where \(\tau\) is the torque, \(K\) is the torsional constant for wire and equal to \(\pi G r^{4} / 2 l, G\) is the modulus of rigidity for copper and equal to \(45 \mathrm{GPa}, l\) is the length, and \(\theta\) is the angular displacement from equilibrium. Calculate the period of the harmonic motion.

A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude \(A\) and period \(T\). How long does it take for the object to travel a total distance of \(6 \mathrm{~A}\) ? A. \(T / 2\) D. \(3 T / 2\) B. \(3 T / 4\) E. \(2 T\) C. \(T\)
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