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A glider is oscillating in 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 \(x=\pm A_{1} / 4 ;(\text { e) potential }\) and kinetic energies at \(x=\pm A_{1} / 4 ?\)

Short Answer

Expert verified
The period, frequency, and angular frequency remain the same. The total mechanical energy becomes 1/4 of the original; the maximum speed halved. Speed at \(\pm A_1/4\) and energies decrease.

Step by step solution

01

Period, Frequency, and Angular Frequency

For an object undergoing simple harmonic motion (SHM), the period \( T \), frequency \( f \), and angular frequency \( \omega \) are independent of amplitude. So, slowing the glider and halving its amplitude does not affect these parameters.- **Period (\( T \))** is given by \( T = 2\pi\sqrt{\frac{m}{k}} \) and remains constant.- **Frequency (\( f \))** is \( f = \frac{1}{T} \) and remains constant.- **Angular frequency (\( \omega \))** is \( \omega = 2\pi f = \sqrt{\frac{k}{m}} \) and remains constant.
02

Total Mechanical Energy

The total mechanical energy \( E \) in SHM is given by \( E = \frac{1}{2}kA^2 \), where \( A \) is the amplitude. If the amplitude is halved \(( A_2 = \frac{A_1}{2})\), the new energy \( E_2 \) becomes:\[ E_2 = \frac{1}{2}k\left(\frac{A_1}{2}\right)^2 = \frac{1}{4} \times \frac{1}{2}kA_1^2 = \frac{1}{4}E_1 \]Thus, the total mechanical energy is reduced to one-fourth of its original value.
03

Maximum Speed

The maximum speed \( v_{max} \) in SHM is given by \( v_{max} = A\omega \). When the amplitude is halved, the new maximum speed becomes:\[ v_{max, new} = \frac{A_1}{2}\omega = \frac{1}{2}v_{max, old} \]The maximum speed is halved.
04

Speed at \(x = \pm A_1/4\)

The speed at displacement \( x = \pm \frac{A_1}{4} \) for a system in SHM can be determined by the equation \( v = \omega\sqrt{A^2-x^2} \). For the new amplitude, this gives:\[ v_{new} = \omega\sqrt{\left(\frac{A_1}{2}\right)^2-\left(\frac{A_1}{4}\right)^2} = \omega\sqrt{\frac{A_1^2}{4} - \frac{A_1^2}{16}} = \frac{\omega A_1\sqrt{3}}{4}\]The speed reduces, and it can be expressed as \( \frac{v_{old}}{2}\sqrt{\frac{3}{2}} \), assuming \( v_{old} \) is the speed at \( \pm \frac{A_1}{4} \) before halving the amplitude.
05

Potential and Kinetic Energies at \(x = \pm A_1/4\)

In SHM, the potential energy at any displacement \( x \) is given by \( U = \frac{1}{2}kx^2 \). For \( x = \pm \frac{A_1}{4} \):- **Potential Energy (\( U \))**:\[ U = \frac{1}{2}k\left(\frac{A_1}{4}\right)^2 = \frac{1}{32}kA_1^2 \] for the new amplitude.- **Kinetic Energy (\( K \))** can be obtained by \( K = E - U \):\[ K = \frac{1}{4}E_1 - \frac{1}{32}kA_1^2 = \frac{7}{32}kA_1^2 \]Both energies are altered due to the amplitude change, with each decreasing correspondingly.

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Key Concepts

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

Amplitude
In simple harmonic motion (SHM), the term **amplitude** refers to the maximum extent of displacement from the equilibrium position. It indicates how far the object swings or moves in either direction during its motion. The amplitude is a crucial parameter as it affects other aspects like speed but does not affect the time-related properties such as period and frequency.

Even if you reduce the amplitude, say by halving it, several aspects like period, frequency, and angular frequency remain unchanged. However, other characteristics like maximum speed and mechanical energy are directly affected by changes in amplitude.
Period
The **period** in SHM is the time it takes for an object to complete one full cycle of motion. It is symbolized by the letter \( T \) and is calculated using the formula: - \( T = 2\pi\sqrt{\frac{m}{k}} \) Here, \( m \) is the mass and \( k \) is the spring constant.

Notably, the period is independent of the amplitude in SHM. This independence means that even if you change the amplitude, even considerably, the period remains the same. The glider, in our exercise, has a period unaffected by reducing the amplitude.
Mechanical Energy
**Mechanical energy** in SHM is composed of kinetic and potential energy. It is the conserved energy of the system under ideal conditions. The total mechanical energy \( E \) in SHM can be expressed as: - \( E = \frac{1}{2}kA^2 \) where \( A \) represents the amplitude.

When the amplitude is halved, the mechanical energy decreases greatly, specifically to a quarter of its original value. This significant reduction stems from the square relationship in the energy equation, showcasing how mechanics here are directly sensitive to amplitude changes.
Frequency
**Frequency** refers to how many oscillations or cycles occur in one second. It is denoted by the symbol \( f \) and is the reciprocal of the period: - \( f = \frac{1}{T} \) Since the period remains constant regardless of amplitude changes, frequency also stays the same. Whether the oscillation is large or small, as evidenced by the glider, frequency represents the temporal consistency of SHM that doesn't bow to amplitude shifts.

Frequency reflects the rhythmic, repeatable nature characteristic of systems in SHM, similar to beats in music, maintaining a steady pace even as other parameters adjust.
Angular Frequency
**Angular frequency** is expressed as \( \omega \) and measures how fast the object moves through its cycle in terms of the angle covered per unit time. It is connected to both the period and frequency by these formulas: - \( \omega = 2\pi f = \sqrt{\frac{k}{m}} \) Given the relations of \( \omega \) to other constants, it too does not alter with changes in amplitude, maintaining stability in characteristic motion speed.

Through angular frequency, we understand how SHM translates linear motion into rotational analogies, reflecting an elegant uniformity. In our context, even as amplitude changes, the angular narrative remains consistent, showcasing uniformity in rotational dynamics.

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Most popular questions from this chapter

Jerk. A guitar string vibrates at a frequency of 440 \(\mathrm{Hz}\) . A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

A \(1.80-\mathrm{kg}\) monkey wrench is pivoted 0.250 \(\mathrm{m}\) from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

A \(2.00-\mathrm{kg}\) bucket containing 10.0 \(\mathrm{kg}\) of water is hanging from a vertical ideal spring of force constant 125 \(\mathrm{N} / \mathrm{m}\) 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 building in San Francisco has light fixtures consisting of small \(2.35-\mathrm{kg}\) bulbs with shades hanging from the ceiling at the end of light thin cords \(1.50 \mathrm{~m}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

On the planet Newtonia, a simple pendulum having a bob with mass 1.25 \(\mathrm{kg}\) and a length of 185.0 \(\mathrm{cm}\) takes 1.42 \(\mathrm{s}\) , when released from rest, to swing through an angle of \(12.5^{\circ},\) where it again has zero speed. The circumference of Newtonia is measured to be \(51,400 \mathrm{km}\) . What is the mass of the planet Newtonia?

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