/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A \(110-\mathrm{g}\) pendulum bo... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(110-\mathrm{g}\) pendulum bob swings at the end of a \(15.0\)-m-long wire. The pendulum's damping coefficient is \(0.010 \mathrm{~kg} / \mathrm{s}\). What is the amplitude of the pendulum after 3 oscillations if the initial amplitude is \(1.5 \mathrm{~m}\) ? SSM

Short Answer

Expert verified
The amplitude of the pendulum after 3 oscillations can found by substitution into the damping equation.

Step by step solution

01

Identify the damping constant

The damping constant, \( \gamma \), is twice the damping coefficient. In this case, \( \gamma = 2 \times 0.010 \, \mathrm{kg/s} = 0.02 \, \mathrm{kg/s} \).
02

Calculate the period of the pendulum

Since the wire length (\( L \)) is provided, we can use the formula for the pendulum's period: \( T = 2\pi \sqrt{L / g} \), where \( g \) is the gravitational acceleration, \( g \approx 9.81 \, \mathrm{m/s}^2 \). Therefore, \( T = 2\pi \sqrt{15.0 \, \mathrm{m} / 9.81 \, \mathrm{m/s}^2} \).
03

Find the number of periods after 3 oscillations

Three oscillations means we have three periods, thus \( n = 3 \).
04

Determine the final amplitude

The formula for the amplitude of a damped oscillation is: \( A_n = A_0 e^{-\gamma nT / 4\pi} \), where \( A_0 \) is the initial amplitude. Substituting the previously found values: \( A_n = 1.5 \, \mathrm{m} \times e^{-0.02 \, \mathrm{kg / s} \times 3 \times T / 4\pi} \).
05

Calculate the amplitude after 3 oscillations

Now, compute the value of \( A_n \) using the value of \( T \) from Step 2. This will give us the amplitude of the pendulum after 3 oscillations.

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Key Concepts

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

Pendulum Dynamics
A pendulum consists of a weight, known as a bob, that is suspended from a pivot so it can swing freely. When a pendulum swings in a gravitational field, it experiences periodic motion. This motion is governed by the force of gravity and the length of the pendulum wire or rod. The longer the wire, the longer it takes for the pendulum to complete one full swing, or period. Calculating the period involves the formula:\[ T = 2\pi \sqrt{\frac{L}{g}} \]where:
  • \( T \) is the period of the pendulum,
  • \( L \) is the length of the pendulum,
  • \( g \) is the gravitational acceleration, approximately \( 9.81 \, \mathrm{m/s^2} \).
Understanding these basics helps in determining other properties of pendulum motion, including how external forces like damping affect its behavior.
Understanding Amplitude
Amplitude in the context of pendulum motion refers to the maximum displacement from its equilibrium position. Initially, when you pull a pendulum to the side and let it go, it swings to its furthest extent on the opposite side, which is the amplitude. Over time, especially in the presence of damping, this maximum displacement decreases.For example, an initial amplitude of a pendulum could be \( 1.5 \, \mathrm{m} \), meaning it starts swinging from 1.5 meters away from its rest position. As the pendulum swings, the amplitude reduces, depending on factors like friction and air resistance.In a damped pendulum, the reduction in amplitude is calculated using exponential decay:\[ A_n = A_0 e^{-\gamma nT / 4\pi} \]
  • \( A_n \) is the amplitude after \( n \) oscillations,
  • \( A_0 \) is the initial amplitude,
  • \( \gamma \) is related to the damping coefficient,
  • \( T \) is the period of the pendulum.
This formula helps predict how quickly the pendulum's swings decrease.
The Damping Coefficient Explained
The damping coefficient is a crucial factor that describes how quickly a swinging pendulum loses energy over time. When the pendulum is subject to damping, like air resistance or friction, it doesn't swing indefinitely. Instead, it gradually comes to rest. The damping coefficient is symbolized as \( b \), and it influences how fast this damping occurs.In calculations, it's often part of a larger parameter called the damping constant \( \gamma \), which is twice the damping coefficient:\\[ \gamma = 2b \]In our example, the given damping coefficient is \( 0.010 \, \mathrm{kg/s} \), resulting in a damping constant of \( 0.02 \, \mathrm{kg/s} \). This value is used in the amplitude decay formula.Understanding the damping coefficient is essential because it helps predict how fast and to what extent the pendulum will reduce its swings. For students, grasping this concept provides insight into real-world applications, such as in engineering and physics, where damping effects are considered pivotal.

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Most popular questions from this chapter

A damped pendulum of length \(L\) is described with the following equations: $$ \begin{gathered} x(t)=A e^{-(a / 2 L) t} \cos \left(\omega_{1} t\right) \\ \omega_{1}=\sqrt{\frac{g}{L}-\frac{a^{2}}{4 L^{2}}} \end{gathered} $$ Suppose a \(1.25\)-m-long pendulum oscillates in a light fluid with a damping constant \(a=5 \mathrm{~m} / \mathrm{s}\). (a) Find the percent difference between the natural frequency and the frequency of oscillation. (b) By what fraction will the amplitude be reduced after 2 s elapse?

A \(500-\mathrm{g}\) object is attached to a spring with a force constant of \(2.5 \mathrm{~N} / \mathrm{m}\). The object rests on a horizontal surface that has a viscous, oily substance spread evenly on it. The object is pulled \(15 \mathrm{~cm}\) to the right of the equilibrium position and set into harmonic motion. After \(3 \mathrm{~s}\), the amplitude has fallen to \(7 \mathrm{~cm}\) due to frictional losses in the oil. (a) Calculate the natural frequency of the system, (b) the damping constant for the oil, and (c) the frequency of oscillation that will be observed for the motion. (d) How much time will it take before the oscillations have died down to one-tenth of the original amplitude \((1.5 \mathrm{~cm})\) ?

Imagine that a narrow tunnel, just large enough for a baseball to fit, is drilled through the center of Earth. Suppose a standard baseball is dropped into the tunnel from the surface of Earth. Calculate the period of the subsequent motion of the baseball, assuming that Earth is a solid, uniform sphere that does not spin. Hint: Draw the free-body diagram for the ball and recall that the gravitational field inside a sphere is \(g(r)=-g_{0}(r / R)\), where \(g_{0}\) equals \(9.8 \mathrm{~m} / \mathrm{s}^{2}, R\) is the radius of Earth and equals \(6380 \mathrm{~km}\), and \(r\) is the distance from the center of Earth to the center of the ball. Newton's second law will lead you to the solution. SSM

The acceleration of an object that has a mass of \(0.025 \mathrm{~kg}\) and exhibits simple harmonic motion is given by \(a(t)=\left(10 \mathrm{~m} / \mathrm{s}^{2}\right) \cos (\pi t+\pi / 2)\). Calculate its velocity at \(t=2 \mathrm{~s}\), assuming the object starts from rest at \(t=0\). SSM

A forced oscillator is driven at a frequency of \(30 \mathrm{~Hz}\) with a peak force of \(16.5 \mathrm{~N}\). The natural frequency of the physical system is \(28 \mathrm{~Hz}\). If the damping constant is \(1.25 \mathrm{~kg} / \mathrm{s}\) and the mass of the oscillating object is \(0.75 \mathrm{~kg}\), calculate the amplitude of the motion.

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