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Chapter 14: Problem 59

A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=\) \(1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equi. librium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0 .\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0\), the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.

Short Answer

Expert verified
Question: Write the equation of motion for the spring-mass system described in the following cases: a) The mass is initially moving to the right with a speed of \(1.00~m/s\) and passing its equilibrium position. b) The mass is initially \(0.50~m\) to the right of its equilibrium position and moving to the right with a speed of \(1.00~m/s\). Answer: a) \(x(t) = 1 \cdot \cos(1 \cdot t + \frac{\pi}{2})\) b) \(x(t) \approx 1.12\cos(1 \cdot t + 1.05)\)

Step by step solution

01

Calculate the angular frequency.

The angular frequency, \(\omega\), is related to the spring constant and the mass. It is given by the formula: \(\omega=\sqrt{\frac{k}{m}}\) Given, \(k = 1.00~N/m\) and \(m= 1.00~kg\), we can calculate the angular frequency: \(\omega = \sqrt{\frac{1}{1}}\) \(\omega = 1~rad/s\)
02

Identify the form of the equation of motion.

Since the system experiences simple harmonic motion, we can represent the equation of motion in the following form: \(x(t) = A\cos(\omega t+\varphi)\) where \(x(t)\) is the position of the mass at time \(t\), \(A\) is the amplitude, \(\omega\) is the angular frequency (which we calculated in Step 1), and \(\varphi\) is the phase.
03

Solve for the amplitude and phase for case a).

For case a), we are given that at time \(t = 0\), the mass is moving to the right with a speed of \(1.00~m/s\) and passing its equilibrium position (\(x = 0\)). Using these initial conditions, we can write: \(x(0) = A\cos(\omega \cdot 0 + \varphi) = 0\) and \(v(0) = -A \omega \sin(\omega \cdot 0 + \varphi) = 1.00~m/s\) From the first equation, we have: \(\cos(\varphi) = 0\) This means that the phase, \(\varphi\), is either \(\frac{\pi}{2}\) or \(-\frac{\pi}{2}\). We can use the second equation to find the amplitude and the correct value of the phase: \(A = \frac{v(0)}{\omega \sin(\varphi)}\) Using \(\varphi = \frac{\pi}{2}\), \(A = \frac{1}{1 \cdot 1} = 1~m\)
04

Write the equation of motion for case a).

Now that we have the amplitude and phase for case a), we can write the equation of motion: \(x(t) = 1 \cdot \cos(1 \cdot t + \frac{\pi}{2})\)
05

Solve for the amplitude and phase for case b).

For case b), we are given that at time \(t = 0\), the mass is at \(x = 0.50~m\) and moving to the right with a speed of \(1.00~m/s\). Using these initial conditions, we can write: \(x(0) = A\cos(\omega \cdot 0 + \varphi) = 0.50~m\) and \(v(0) = -A \omega \sin(\omega \cdot 0 + \varphi) = 1.00~m/s\) From the second equation, we can solve for the amplitude and phase: \(A = \frac{v(0)}{\omega \sin(\varphi)}\) Using \(x(0)\) and \(\varphi = \arccos(\frac{x(0)}{A})\), \(A = \sqrt{x(0)^2 + (\frac{v(0)}{\omega})^2}\) \(A = \sqrt{(0.50)^2 + (\frac{1}{1})^2}\) \(A = \sqrt{1.25} \approx 1.12~m\) Now, we can find the phase: \(\varphi = \arccos(\frac{x(0)}{A})\) \(\varphi = \arccos(\frac{0.50}{1.12})\) \(\varphi \approx 1.05~rad\)
06

Write the equation of motion for case b).

Now that we have the amplitude and phase for case b), we can write the equation of motion: \(x(t) \approx 1.12\cos(1 \cdot t + 1.05)\)

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Key Concepts

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

Spring-Mass System
In a spring-mass system, an object with mass is attached to a spring that can compress and extend. This setup is a classic example of simple harmonic motion, where the mass moves back and forth around an equilibrium position. The force exerted by the spring follows Hooke's Law, which states that the force is directly proportional to the displacement of the spring from its equilibrium position: \[ F = -kx \]
  • **Spring Constant (k)**: This is a measure of the stiffness of the spring. A higher spring constant means a stiffer spring.
  • **Mass (m)**: This is the object's mass attached to the spring. Changes in mass will influence the motion dynamics.
The spring-mass system without any damping or external forces (ideal conditions) leads to simple harmonic motion. The mass oscillates at a natural frequency determined by the spring constant and its own mass.
Angular Frequency
Angular frequency plays a crucial role in describing the dynamics of simple harmonic motion. It tells us how quickly the system oscillates. It is given by the formula:\[ \omega = \sqrt{\frac{k}{m}} \]
  • Where **\(\omega\)** is the angular frequency measured in radians per second.
  • **k** is the spring constant, which defines the spring's stiffness.
  • **m** is the mass of the object attached to the spring.
This formula shows that a stiffer spring (higher k) or a smaller mass (lower m) results in a higher angular frequency, indicating faster oscillations. In the given problem, with \(k = 1.00 \) and \(m = 1.00 \) kg, the angular frequency comes out to be \(1\) rad/s, meaning the mass completes its cycle every \(2\pi\) seconds.
Equation of Motion
The equation of motion for a spring-mass system provides a mathematical representation of the mass's position as a function of time. It is generally expressed as:\[ x(t) = A \cos(\omega t + \varphi) \]
  • **x(t)**: The position of the mass at any time \(t\).
  • **A**: The amplitude, representing the maximum displacement from equilibrium.
  • **\(\omega\)**: Angular frequency, dictating how quickly the mass oscillates.
  • **\(\varphi\)**: The phase angle, which adjusts the function to match initial conditions.
For case (a) in the exercise, the initial conditions were such that the mass passed through equilibrium with a specific velocity, leading to an equation: \[ x(t) = 1 \cdot \cos(1 \cdot t + \frac{\pi}{2}) \]For case (b), where the mass had an initial displacement and a velocity, the equation adjusted to:\[ x(t) \approx 1.12 \cos(1 \cdot t + 1.05) \]These equations capture the essence of the motion under the specified conditions and are foundational for understanding oscillatory systems like spring-mass units.

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Most popular questions from this chapter

In a lab, a student measures the unstretched length of a spring as \(11.2 \mathrm{~cm}\). When a 100.0 - g mass is hung from the spring, its length is \(20.7 \mathrm{~cm}\). The mass-spring system is set into oscillatory motion, and the student obscrves that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles. a) Calculate the period of oscillation for this system, assuming no damping. b) If the student can measure the period to the nearest \(0.05 \mathrm{~s}\). will she be able to detect the difference between the period with no damping and the period with damping?

If you kick a harmonic oscillator sharply, you impart to it an initial velocity but no initial displacement. For a weakly damped oscillator with mass \(m\), spring constant \(k\). and damping force \(F_{y}=-b v,\) find \(x(t),\) if the total impulse delivered by the kick is \(J_{0}\).

A mass \(m\) is attached to a spring with a spring constant of \(k\) and set into simple harmonic motion. When the mass has half of its maximum kinetic energy, how far away from its equilibrium position is it, expressed as a fraction of its maximum displacement?

A spring is hanging from the ceiling with a mass attached to it. The mass is pulled downward, causing it to oscillate vertically with simple harmonic motion. Which of the following will increase the frequency of oscillation? a) adding a second, identical spring with one end attached to the mass and the other to the ceiling b) adding a second, identical spring with one end attached to the mass and the other to the floor c) increasing the mass d) adding both springs, as described in (a) and (b)

A physical pendulum consists of a uniform rod of mass \(M\) and length \(L\) The pendulum is pivoted at a point that is a distance \(x\) from the center of the rod, so the period for oscillation of the pendulum depends on \(x: T(x)\). a) What value of \(x\) gives the maximum value for \(T ?\) b) What value of \(x\) gives the minimum value for \(T ?\)
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