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A glider is oscillating in \(\mathrm{SHM}\) on an air track with an amplitude \(A_{1 .}\) You slow it so that its amplitude is halved. What happens to its (a) period, frequency, and angular frequency; (b) total mechanical energy; (c) maximum speed; (d) speed at \(x=\pm A_{1} / 4 ;\) (e) potential and kinetic energies at \(x=\pm A_{1} / 4 ?\)

Short Answer

Expert verified
(a) Period, frequency, and angular frequency remain unchanged; (b) Total energy becomes one-quarter of initial; (c) Maximum speed is halved; (d) Speed at \( x = \pm \frac{A_1}{4} \) is \( \frac{\omega A_1}{2\sqrt{2}} \); (e) Use energy equations to find potential and kinetic energies.

Step by step solution

01

Understanding Initial Conditions

Initially, the glider has an amplitude \( A_1 \) and it undergoes simple harmonic motion (SHM). Key formulas include the period \( T = 2\pi \sqrt{\frac{m}{k}} \) and angular frequency \( \omega = \sqrt{\frac{k}{m}} \), which do not depend on amplitude.
02

Halving the Amplitude

The amplitude is reduced to \( \frac{A_1}{2} \). We need to determine the impact of this change on the glider's motion properties.
03

Effects on Period and Frequency (a)

The period \( T \) and frequency \( f \) of SHM are independent of amplitude. Therefore, halving the amplitude does not affect the period, frequency, or angular frequency. Thus, \( T \), \( f \), and \( \omega \) remain unchanged.
04

Total Mechanical Energy (b)

The total mechanical energy in SHM is given by \( E = \frac{1}{2}kA^2 \). If the amplitude is halved, the energy becomes \( E = \frac{1}{2}k\left(\frac{A_1}{2}\right)^2 = \frac{1}{4}E_1 \). The total mechanical energy is now one-quarter of the initial value.
05

Maximum Speed (c)

The maximum speed \( v_{max} \) is given by \( v_{max} = A\omega \). With amplitude halved, the maximum speed becomes \( v_{max} = \frac{A_1}{2}\omega \), or half of the initial maximum speed.
06

Speed at \( x = \pm \frac{A_1}{4} \) (d)

Using conservation of energy, the speed \( v \) at position \( x = \pm \frac{A_1}{4} \) is \( v = \omega\sqrt{A^2 - x^2} \). When \( A = \frac{A_1}{2} \), the speed becomes \( v = \omega\sqrt{\left(\frac{A_1}{2}\right)^2 - \left(\frac{A_1}{4}\right)^2} \), simplifying to \( v = \frac{\omega A_1}{2\sqrt{2}} \).
07

Potential and Kinetic Energies at \( x = \pm \frac{A_1}{4} \) (e)

The potential energy at \( x = \pm \frac{A_1}{4} \) is \( U = \frac{1}{2}kx^2 \). At \( x = \pm \frac{A_1}{4} \), \( U = \frac{1}{2}k\left(\frac{A_1}{4}\right)^2 \). The kinetic energy \( K \) is given by \( E - U \), where \( E \) is the total mechanical energy after the amplitude change.

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Key Concepts

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

Amplitude
Amplitude is a crucial part of simple harmonic motion (SHM). It represents the maximum distance that a moving object in SHM travels from its equilibrium rest position. For example, if a glider oscillates back and forth on a track, the amplitude is its farthest distance from the center.
In the given problem, the glider’s amplitude is initially denoted as \( A_1 \). When the amplitude is halved, it means the glider can only reach half of the original maximum displacement. This change in amplitude impacts other motion aspects, such as energy and speed, but not the period or frequency.
  • **Original Amplitude**: Maximum displacement \( A_1 \)
  • **New Amplitude**: Reduced to \( \frac{A_1}{2} \)
Period and Frequency
In SHM, the period \( T \) is the time taken to complete one full cycle of motion. The frequency \( f \), on the other hand, is how many cycles occur in a unit time, usually per second, and is the reciprocal of the period \( f = \frac{1}{T} \). Angular frequency \( \omega \) relates to how fast the object is oscillating in radian measure.
Importantly, the period and frequency of SHM are independent of the amplitude. This means halving the amplitude of the glider does not affect these measures at all. These properties depend on the mass of the object and the spring constant of the system, following the formulas \( T = 2\pi \sqrt{\frac{m}{k}} \) and \( \omega = \sqrt{\frac{k}{m}} \).
  • **Period \( T \)**: Unchanged
  • **Frequency \( f \)**: Unchanged
  • **Angular Frequency \( \omega \)**: Unchanged
Mechanical Energy
Mechanical energy in SHM is made up of kinetic energy and potential energy. The total mechanical energy \( E \) of a system in SHM is given by the formula \( E = \frac{1}{2}kA^2 \), where \( k \) is the spring constant and \( A \) is the amplitude. This indicates that mechanical energy is directly proportional to the square of the amplitude.
Therefore, when the amplitude is halved to \( \frac{A_1}{2} \), the total mechanical energy becomes one-quarter of the original. This occurs because \[ E_{new} = \frac{1}{2}k\left(\frac{A_1}{2}\right)^2 = \frac{1}{4}E_1 \],meaning the energy stored in the motion decreases substantially.
  • **Initial Energy**: \( E_1 = \frac{1}{2}kA_1^2 \)
  • **New Energy**: \( E_{new} = \frac{1}{4}E_1 \)
Conservation of Energy
In SHM, the principle of conservation of energy is a key concept. It states that energy within a closed system remains constant over time. This means even if kinetic and potential energy vary throughout the motion, their sum, which is the total mechanical energy, stays the same.
With the halved amplitude, while both the potential and kinetic energies will alter their individual values at any given position, the sum reflects the reduced total mechanical energy already discussed. For instance, when the glider is at \( x = \pm \frac{A_1}{4} \), you can find potential energy \( U \) and kinetic energy \( K \) using energy expressions.
  • **Potential Energy \( U \)** at \( x = \pm \frac{A_1}{4} \): \( U = \frac{1}{2}k\left(\frac{A_1}{4}\right)^2 \)
  • **Kinetic Energy \( K \)**: \( K = E_{new} - U \)
Conservation of energy dictates that these will dynamically add up to the new total mechanical energy throughout the oscillation.

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Most popular questions from this chapter

An object is undergoing SHM with period 0.300 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 \(x=-1.50 \mathrm{cm} .\)

A 2.00 -kg, frictionless block is attached to an ideal spring with force constant 300 \(\mathrm{N} / \mathrm{m} .\) At \(t=0\) the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 \(\mathrm{m} / \mathrm{s} .\) Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

CP SHM of a Floating Object. An object with height \(h,\) mass \(M,\) and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\) (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7\()\) .

You pull a simple pendulum 0.240 m long to the side through an angle of \(3.50^{\circ}\) and release it. (a) How much time does it take the pendulum bob to reach its highest speed? (b) How much time does it take if the pendulum is released at an angle of \(1.75^{\circ}\) instead of \(3.50^{\circ} ?\)

A \(175-\) 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?

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