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The initial position, velocity, and acceleration of an object moving in simple harmonic motion are \(x_{i}, v_{i},\) and \(a_{i} ;\) the angular frequency of oscillation is \(\omega .\) (a) Show that the position and velocity of the object for all time can be written as,$$\begin{array}{l}x(t)=x_{i} \cos \omega t+\left(\frac{v_{i}}{\omega}\right) \sin \omega t \\\v(t)=-x_{i} \omega \sin \omega t+v_{i}\cos \omega t\end{array}$$ (b) If the amplitude of the motion is \(A\), show that,$$v^{2}-a x=v_{i}^{2}-a_{i} x_{i}=\omega^{2} A^{2}$$.

Short Answer

Expert verified
The position \(x(t)\) and velocity \(v(t)\) of the object can be written as \(x(t) = x_i cos(\omega t) + \frac{v_i}{\omega}sin(\omega t)\) and \(v(t) = -x_i \omega sin(\omega t) + v_i cos(\omega t)\) respectively. The equation \(v^2 - ax = v_i^2 - a_i x_i = \omega^2 A^2\) holds true.

Step by step solution

01

Determining the Equations of Position and Velocity

By definition, the position \(x(t)\) and velocity \(v(t)\) of the object in Simple Harmonic Motion (SHM) at any time \(t\) can be expressed in terms of initial position \(x_i\), initial velocity \(v_i\), angular frequency \(\omega\), and acceleration \(a_i\). The solutions are obtained using the equations for SHM: \[x(t) = x_i cos(\omega t) + \frac{v_i}{\omega}sin(\omega t)\] \[v(t) = -x_i \omega sin(\omega t) + v_i cos(\omega t)\]
02

Determining the equation for \(v^2 - ax\)

Firstly, square the equation for velocity obtained in step 1, and multiply the position equation by acceleration, \(a\), and then subtract the two. This will give \(v^2 - a x\). It simplifies to: \[v^2 - ax = v_i^2 - a_i x_i\]
03

Including Amplitude in Equation

The amplitude of the motion, \(A\), is given by \[A = sqrt{(x_i^2 + (v_i/\omega)^2)}\]. Squaring this equation gives \[A^2 = x_i^2 + (v_i/\omega)^2\]. By comparing with \(v^2 - ax = v_i^2 - a_i x_i\), it can be concluded that \[v_i^2 - a_i x_i = \omega^2 A^2\]

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Key Concepts

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

Angular Frequency
Angular frequency, denoted by the Greek letter \( \omega \) in physics, is a measure of how quickly an object moves through its cycle in Simple Harmonic Motion (SHM). Imagine a child on a swing; the angular frequency would tell us how fast the swing goes back and forth in a certain amount of time. It's closely related to the period \( T \) of the motion, which is the time it takes to complete one cycle. The relationship between them is \( \omega = 2\pi / T \) where \( 2\pi \) represents a full circle in radians.

Higher angular frequency means that the object oscillates more quickly, hence completing more cycles in a given time frame. In the textbook solution, we see the angular frequency used to describe the position and velocity of an object in SHM with the equations:\begin{array}{l}x(t)=x_{i} \cos \omega t+\left(\frac{v_{i}}{\omega}\right) \sin \omega t \v(t)=-x_{i} \omega \sin \omega t+v_{i}\cos \omega t\end{array}\. These equations imply that the position and velocity are periodic and vary sinusoidally with time, at a rate determined by the angular frequency.
SHM Amplitude
The amplitude in SHM, usually represented by \( A \), refers to the maximum displacement from the equilibrium position. Think of it as the highest point the child's swing reaches from the rest position. The amplitude is a crucial characteristic of any oscillatory system because it demonstrates the energy involved in the motion; larger amplitudes mean more energy is stored in the system.

In the exercise, the amplitude \( A \) is shown to be related to the initial position and velocity, as well as the angular frequency through the equation:\begin{array}{l}A = \sqrt{x_{i}^{2} + \(\frac{v_{i}}{\omega}\)^{2}}\end{array}\. This equation is derived using the Pythagorean theorem in the context of SHM, considering the initial position and velocity as perpendicular components of the motion. It's essential for students to grasp that the amplitude remains constant in a system with no energy loss, irrespective of the position or the velocity at which you measure it.
Position and Velocity in SHM
In Simple Harmonic Motion, position \( x(t) \) and velocity \( v(t) \) are inherently connected; the velocity is the rate of change of position. The formulas given in the exercise solution are fundamental for understanding how the position and velocity change over time in SHM:\begin{array}{l}x(t)=x_{i} \cos \omega t+\left(\frac{v_{i}}{\omega}\right) \sin \omega t \v(t)=-x_{i} \omega\sin \omega t+v_{i}\cos \omega t\end{array}\.

These equations show that both position and velocity oscillate sinusoidally. The \( \cos \) and \( \sin \) functions indicate that the motion is repetitive and periodic. Importantly, when the position reaches its maximum value (amplitude), the velocity is zero because, at the highest point, the object momentarily stops before reversing direction. Conversely, when the object passes through the equilibrium position, its velocity reaches a maximum.

Another point to note from these equations is the phase difference: position and velocity waves are offset by a 90-degree phase, meaning when one is at a maximum, the other is at zero. It's this relationship between position and velocity that underpins the dynamic behavior of any SHM system.

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Most popular questions from this chapter

A \(10.6-\mathrm{kg}\) object oscillates at the end of a vertical spring that has a spring constant of \(2.05 \times 10^{4} \mathrm{N} / \mathrm{m} .\) The effect of air resistance is represented by the damping coefficient \(b=3.00 \mathrm{N} \cdot \mathrm{s} / \mathrm{m} . \quad\) (a) Calculate the frequency of the damped oscillation. (b) By what percentage does the amplitude of the oscillation decrease in each cycle? (c) Find the time interval that elapses while the energy of the system drops to \(5.00 \%\) of its initial value.

A particle of mass \(m\) slides without friction inside a hemispherical bowl of radius \(R\). Show that, if it starts from rest with a small displacement from equilibrium, the particle moves in simple harmonic motion with an angular frequency equal to that of a simple pendulum of length \(R\) That is, \(\omega=\sqrt{g / R}\).

(a) A hanging spring stretches by \(35.0 \mathrm{cm}\) when an object of mass \(450 \mathrm{g}\) is hung on it at rest. In this situation, we define its position as \(x=0 .\) The object is pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. What is its position \(x\) at a time 84.4 s later? (b) What If? A hanging spring stretches by \(35.5 \mathrm{cm}\) when an object of mass \(440 \mathrm{g}\) is hung on it at rest. We define this new position as \(x=0 .\) This object is also pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. Find its position 84.4 s later. (c) Why are the answers to (a) and (b) different by such a large percentage when the data are so similar? Does this circumstance reveal a fundamental difficulty in calculating the future? (d) Find the distance traveled by the vibrating object in part (a). (e) Find the distance traveled by the object in part (b).

A block of unknown mass is attached to a spring with a spring constant of \(6.50 \mathrm{N} / \mathrm{m}\) and undergoes simple harmonic motion with an amplitude of \(10.0 \mathrm{cm} .\) When the block is halfway between its equilibrium position and the end point, its speed is measured to be \(30.0 \mathrm{cm} / \mathrm{s}\). Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block.

A \(0.500-\mathrm{kg}\) object attached to a spring with a force constant of \(8.00 \mathrm{N} / \mathrm{m}\) vibrates in simple harmonic motion with an amplitude of \(10.0 \mathrm{cm} .\) Calculate (a) the maximum value of its speed and acceleration, (b) the speed and acceleration when the object is \(6.00 \mathrm{cm}\) from the equilibrium position, and (c) the time interval required for the object to move from \(x=0\) to \(x=8.00 \mathrm{cm}\).

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