Problem 69 - University Physics with Modern Physics

Question

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 \(x=\pm A_{1} / 4 ?\)

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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.

Step 5: 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.

Step 6: 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 \( v = \frac{\omega A_1}{2\sqrt{2}} \).

Step 7: 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.

Key Concepts

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.