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A \(175 \mathrm{~g}\) glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant \(155 \mathrm{~N} / \mathrm{m}\). At the instant you make measurements on the glider, it is moving at \(0.815 \mathrm{~m} / \mathrm{s}\) and is \(3.00 \mathrm{~cm}\) from its equilibrium point. Use energy conservation to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

Short Answer

Expert verified
The amplitude of the motion is approximately 4.6 cm, the maximum speed of the glider is approximately 0.96 m/s and the angular frequency of the oscillations is approximately 30 rad/s.

Step by step solution

01

Calculate the kinetic energy

We start by calculating the kinetic energy of the glider. This is given by the formula \(\frac{1}{2}mv^2\), where m is the mass and v is the velocity. In this case, \(m = 175 g = 0.175 kg\) and \(v = 0.815 m/s\). So the kinetic energy KE = \( \frac{1}{2}*0.175 kg*(0.815 m/s)^2 \approx 0.0582 J\).
02

Find the potential energy

Next, we compute the potential energy of the glider due to the spring. The expression for potential energy of a spring is \(\frac{1}{2}kx^2\), where k is the spring constant and x is the displacement from equilibrium. Here, \(k = 155 N/m\) and \(x = 3.0 cm = 0.03 m\). Therefore, the potential energy PE = \( \frac{1}{2}*155 N/m*(0.03 m)^2 \approx 0.0698 J\).
03

Calculate the total energy

Since energy is conserved, the total energy, which remains constant, is equal to the sum of the kinetic and potential energies from the first two steps. This gives \(E_{total} = KE + PE = 0.0582 J + 0.0698 J = 0.128 J\).
04

Find the amplitude of oscillation

The amplitude, A, is the maximum displacement of the glider from the equilibrium. At this point, the potential energy is maximum and equals to the total energy, as kinetic energy is zero. So, we use the formula for the potential energy \( PE = \frac{1}{2}kA^2\) to solve for A. We get \( A = \sqrt{\frac{2*E_{total}}{k}} = \sqrt{\frac{2*0.128 J}{155 N/m}} \approx 0.046 m = 4.6 cm\). This is the amplitude of oscillation.
05

Calculate the maximum speed

The maximum speed, corresponds to when the glider is at equilibrium; the kinetic energy is maximum and equals to the total energy (since the potential energy is zero), hence we can solve for the speed using the kinetic energy equation \(E_{total} = \frac{1}{2}mv_{max}^2\), solving for \(v_{max}\), we get \(v_{max} = \sqrt{\frac{2*E_{total}}{m}} = \sqrt{\frac{2*0.128 J}{0.175 kg}} \approx 0.96 m/s\). This is the maximum speed.
06

Find the angular frequency

Lastly, we calculate the angular frequency (\(\omega\)) of the glider. This can be found using the relation \(\omega = \sqrt {\frac{k}{m}}\). Substituting for the given values, we have \(\omega = \sqrt{\frac{155 N/m}{0.175 kg}} \approx 30 rad/s\). This gives us the angular frequency of the oscillations.

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Key Concepts

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

Conservation of Energy in Oscillations
Understanding the conservation of energy in oscillations is crucial for solving problems involving oscillatory motion in physics. It dictates that the total mechanical energy in a system without non-conservative forces, such as friction, remains constant over time.

In the context of an oscillator, like a glider on a frictionless air track attached to an ideal spring, energy conservation implies that at any point in the oscillation, the sum of the kinetic energy (energy of motion) and the potential energy (energy stored in the spring) is constant. This principle can be used to predict numerous properties of the oscillatory motion, such as amplitude and maximum speed, as we see in the given exercise.

To elaborate, when the glider is at the equilibrium point, its potential energy is at a minimum (essentially zero for an ideal spring), and the kinetic energy is at its maximum. Conversely, at the points of maximum displacement (amplitude), the velocity, and thus kinetic energy, is zero, while the potential energy is at its maximum. By knowing the energy at one of these points, you can determine the system's behavior throughout its motion.
Kinetic and Potential Energy of Oscillators
Diving deeper into the energetic aspects of oscillators, the kinetic and potential energy contributions are perpetually interchanging as the system evolves. The kinetic energy, given by the expression \(\frac{1}{2}mv^2\), is at its highest when the object passes through the equilibrium position. Conversely, the potential energy in an ideal spring system, which follows \(\frac{1}{2}kx^2\), peaks when the object is at the furthest point from equilibrium.

This interplay is distinctly observable in the horizontal glider-spring system described in the problem. When the glider momentarily stops at the amplitude of the motion, all the total mechanical energy is stored as potential energy in the spring. Similarly, as the glider passes through the equilibrium point, this potential energy converts into kinetic energy. Through the conservation of energy principle, we can derive that the total mechanical energy is equal to \(E_{total} = KE + PE\), which remains a constant value throughout the motion. Consequently, we can calculate either the potential or kinetic energy at any specific point if we know the total energy of the system.
Angular Frequency of Harmonic Motion
The angular frequency of harmonic motion is a fundamental concept that quantifies the rate at which an object oscillates about its equilibrium position. It is denoted by the symbol \(\omega\) and is measured in radians per second. The equation \(\omega = \sqrt{\frac{k}{m}}\) arises from equating the forces involved in simple harmonic motion.

For a mass-spring system, the spring's restoring force is directly proportional to the displacement according to Hooke's Law \(F = -kx\), and this force is also the mass times acceleration \(F = ma\). When these two forces equate, we can deduce the simple harmonic motion formula which leads to the angular frequency equation.

In our exercise, by knowing the spring constant \(k\) and the mass \(m\) of the glider, we can calculate the glider's angular frequency. This value gives us the speed at which the glider oscillates and is vital for understanding the timing of the oscillation since it is related to the period \(T\) by the relation \(T = \frac{2\pi}{\omega}\). Thus, the angular frequency is inherent to describing the complete motion of the oscillator.

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Most popular questions from this chapter

An object is undergoing SHM with period \(0.300 \mathrm{~s}\) and amplitude \(6.00 \mathrm{~cm} .\) At \(t=0\) the object is instantaneously at rest at \(x=6.00 \mathrm{~cm}\). Calculate the time it takes the object to go from \(x=6.00 \mathrm{~cm}\) to \(x=-1.50 \mathrm{~cm}\)

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A uniform, solid metal disk of mass \(6.50 \mathrm{~kg}\) and diameter \(24.0 \mathrm{~cm}\) hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of \(4.23 \mathrm{~N}\) tangent to the rim of the disk to turn it by \(3.34^{\circ},\) thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

A \(2.00 \mathrm{~kg}\) bucket containing \(10.0 \mathrm{~kg}\) of water is hanging from a vertical ideal spring of force constant \(450 \mathrm{~N} / \mathrm{m}\) and oscillating up and down with an amplitude of \(3.00 \mathrm{~cm}\). Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of \(2.00 \mathrm{~g} / \mathrm{s}\). When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g=3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)

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