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A 10.0 -kg mass is traveling to the right with a speed of 2.00 \(\mathrm{m} / \mathrm{s}\) on a smooth horizontal surface when it collides with and sticks to a second \(10.0-\mathrm{kg}\) mass that is initially at rest but is attached to a light spring with force constant 80.0 \(\mathrm{N} / \mathrm{m}\) . (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

Short Answer

Expert verified
Frequency is 0.318 Hz, amplitude is 0.50 m, period is 3.14 s; time to return is 0.785 s.

Step by step solution

01

Calculate the velocity after the collision

Utilize conservation of momentum: \( m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \).Substitute the values:- \( m_1 = 10 \text{ kg}, v_1 = 2.00 \text{ m/s}\)- \( m_2 = 10 \text{ kg}, v_2 = 0 \text{ m/s}\)The equation becomes:\[ 10 \times 2 + 10 \times 0 = 20 \times v_f \]Solving gives, \( v_f = 1.00 \text{ m/s} \).
02

Determine the angular frequency of oscillation

Use the formula for angular frequency: \( \omega = \sqrt{\frac{k}{m}} \).Where:- Spring constant \( k = 80.0 \text{ N/m} \)- Total mass \( m = 20.0 \text{ kg} \)\[ \omega = \sqrt{\frac{80}{20}} = 2.00 \text{ rad/s} \]
03

Calculate the period of oscillation

Use the relation between period \( T \) and angular frequency: \( T = \frac{2\pi}{\omega} \).Substitute \( \omega = 2.00 \text{ rad/s} \):\[ T = \frac{2\pi}{2} = \pi \text{ seconds} \]
04

Find the amplitude of oscillation

The amplitude \( A \) can be found using the energy method. Initial kinetic energy equals the total potential energy at maximum compression/extension:\[ \frac{1}{2} m v_f^2 = \frac{1}{2} k A^2 \]Substitute \( m = 20 \text{ kg}, v_f = 1 \text{ m/s}, k = 80 \text{ N/m} \):\[ \frac{1}{2} \times 20 \times 1^2 = \frac{1}{2} \times 80 \times A^2 \]Solving gives, \( A = 0.50 \text{ m} \).
05

Calculate time to return to initial phase

The system takes a quarter of the period to return to the positive maximum displacement (which is the point of initial velocity direction change).Thus, time taken is:\[ \text{Time required} = \frac{T}{4} = \frac{\pi}{4} \text{ seconds} \]

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Key Concepts

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

Oscillation Frequency
When two masses collide and stick together, they create an oscillating system, especially if one mass is attached to a spring, as seen in this exercise. To better understand the oscillation, we calculate its frequency. Oscillation frequency is how often the system goes back and forth in one second. It's connected to the angular frequency, denoted as \(\omega\), which is given by the formula:
  • \(\omega = \sqrt{\frac{k}{m}}\)
Here, \(k\) represents the spring constant (80 N/m), and \(m\) is the combined mass (20 kg). After doing the math: \(\omega = 2.00\) rad/s.

The frequency \(f\), measured in hertz, can be found using \(f = \frac{\omega}{2\pi}\). This gives us a value that tells us how often the spring goes through a complete cycle. Understanding this helps anticipate how quickly oscillations occur after the collision. For instance, with \(\omega = 2.00\) rad/s in this scenario, the cycle repeats approximately every 3.14 seconds, equating to a frequency of around 0.318 Hz.
Kinetic Energy
Kinetic energy, often symbolized as KE, is the energy an object possesses due to its motion. In this context, the system's kinetic energy right after the collision is crucial for determining the oscillation's amplitude. Before any oscillation occurs, the kinetic energy of the combined masses right after collision is calculated using:
  • \( KE = \frac{1}{2}mv_f^2 \)
Given that \(m\) stands for the system's total mass (20 kg) and \(v_f\) is the velocity (1 m/s), the kinetic energy is \(10\) Joules.

This energy is transferred to the spring and converted into potential energy at the oscillation's maximum displacement (amplitude). By using energy conservation:
  • \(\frac{1}{2}mv_f^2 = \frac{1}{2}kA^2\)
we find the amplitude, \(A = 0.50\) m. This ensures that we understand how far the spring stretches or compresses at its peak energy state, giving a tangible measure of motion range.
Spring Constant
The spring constant, usually denoted by \(k\), is a measure of a spring's stiffness. It plays a major role in determining how the system behaves after the collision. In our example, the spring constant is specified as 80 N/m, indicating a relatively stiff spring.

The spring constant affects both the oscillation frequency and the system's energy dynamics. It is present in both the angular frequency formula \(\omega = \sqrt{\frac{k}{m}}\) and the amplitude calculation from the kinetic and potential energy relation. A larger \(k\) means the spring is harder to compress or stretch, leading to faster oscillations. Conversely, a smaller \(k\) would mean more flexible oscillations, with cycles taking longer.

In practical terms, understanding \(k\) helps predict how the oscillating masses will behave once they've gained velocity and attached to a spring. It allows us to anticipate physical behaviors such as the period of oscillation (here, \(\pi\) seconds for a full cycle), which is critical when designing systems like vehicle suspension or measuring equipment where oscillations serve a functional purpose.

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Most popular questions from this chapter

A uniform rod of length \(L\) oscillates through small angles about a point a distance \(x\) from its center. (a) Prove that its angular frequency is \(\sqrt{g x /\left[\left(L^{2} / 12\right)+x^{2}\right] .(b)}\) Show that its maximum angular frequency occurs when \(x=L / \sqrt{12}\) . (c) What is the length of the rod if the maximum angular frequency is 2\(\pi \mathrm{rad} / \mathrm{s} ?\)

A mass \(m\) is attached to one end of a massess spring with a force constant \(k\) and an unstretched length \(l_{0}\) . The other end of the spring is free to turn about a nail driven into a frictionless, horizontal surface (Fig. 13.44\() .\) The mass is made to revolve in a circle with an angular frequency of revolution \(\omega^{\prime}\) , (a) Calculate the length \(l\) of the spring as a function of \(\omega^{\prime} .\) (b) What happens to the result in part (a) when \(\omega^{\prime}\) approaches the natural frequency \(\omega=\sqrt{k} / m\) of the mass-spring system? (ff your result bothers you, remember that massless springs and frictionless surfaces don't exist as such, but are only approximate descriptions of real springs and surfaces. Also, Hooke's law is only an approximation of the way real springs behave; the greater the elongation of the spring, the greater the deviation from Hooke's law.)

When a body of unknown mass is attached to an ideal spring with force constant \(120 \mathrm{N} / \mathrm{m},\) it is found to vibrate with a frequency of 6.00 Ha. Find (a) the period of the motion; \((b)\) the angular frequency; (c) the mass of the body.

An object with mass 0.200 \(\mathrm{kg}\) is acted on by an elastic restoring force with force constant 10.0 \(\mathrm{N} / \mathrm{m}\) . (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-0.300 \mathrm{m}\) to \(+0.300 \mathrm{m} .\) On your graph, let \(1 \mathrm{cm}=0.05 \mathrm{J}\) vertically and \(1 \mathrm{cm}=0.05 \mathrm{m}\) borizontally. The object is set into oscillation withan initial potential energy of 0.140 \(\mathrm{J}\) and an initial kinetic energy of 0.060 \(\mathrm{J}\) . Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one-half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 \(\mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\) You twist the part a small amount about this axis and let it go, timing 125 oscillations in 265 \(\mathrm{s}\) . What is the moment of inertia you want to find?

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