Chapter 12: Problem 4
Explain the difference between a simple pendulum and a physical pendulum.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 12: Problem 4
Explain the difference between a simple pendulum and a physical pendulum.
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
An oscillating system has a natural frequency of \(50 \mathrm{rad} / \mathrm{s}\). The damping coefficient is \(2.0 \mathrm{~kg} / \mathrm{s}\). The system is driven by a force \(F(t)=(100 \mathrm{~N}) \cos ((50 \mathrm{rad} / \mathrm{s}) t)\). What is the amplitude of the oscillations? SSM
You can double the maximum speed of a simple harmonic oscillator by A. Doubling the amplitude B. Reducing the mass to one-fourth its original value C. Increasing the spring constant to four times its original value D. All of above E. None of above
A \(60\)-cm-long thin rod of copper has a radius \(r\) of \(0.4 \mathrm{~cm}\) (the density of copper is \(8.92 \mathrm{~g} / \mathrm{cm}^{3}\) ). The rod is suspended from a thin wire that is welded to the exact center of the copper rod. The wire is also made of copper and has a length of \(20 \mathrm{~cm}\) and a cross-sectional diameter of \(1 \mathrm{~mm}\). The rod is displaced from the equilibrium position and the torque on the thin wire causes it to twist the rod back and forth in harmonic motion. This is a torsion pendulum (Figure 12-33). The torque acts on the rod according to following equation: $$ \tau=-K \theta $$ where \(\tau\) is the torque, \(K\) is the torsional constant for wire and equal to \(\pi G r^{4} / 2 l, G\) is the modulus of rigidity for copper and equal to \(45 \mathrm{GPa}, l\) is the length, and \(\theta\) is the angular displacement from equilibrium. Calculate the period of the harmonic motion.
A simple harmonic oscillator completes 1250 cycles in \(20 \mathrm{~min}\). Calculate (a) the period and (b) the frequency of the motion.
A force is measured with a force sensor at the times listed in the following table. (a) Make a plot of force versus time and determine if the force obeys simple harmonic motion. (b) If it is simple harmonic motion, determine the period of the motion. $$ \begin{array}{lr} t(\mathbf{s}) & \boldsymbol{F}(\mathrm{N}) \\ \hline 0 & -20 \\ 0.1 & -10 \\ 0.2 & 0 \\ 0.3 & +10 \\ 0.4 & +20 \\ 0.5 & +10 \\ 0.6 & 0 \\ 0.7 & -10 \\ 0.8 & -20 \\ 0.9 & -10 \\ 1.0 & 0 \\ 1.1 & +10 \\ 1.2 & +20 \end{array} $$ $$ \begin{array}{lr} t(\mathbf{s}) & \boldsymbol{F}(\mathbf{N}) \\ \hline 1.3 & +10 \\ 1.4 & 0 \\ 1.5 & -10 \\ 1.6 & -20 \\ 1.7 & -10 \\ 1.8 & 0 \\ 1.9 & +10 \\ 2.0 & +20 \\ 2.1 & +10 \\ 2.2 & 0 \\ 2.3 & -10 \\ 2.4 & -20 \\ 2.5 & -10 \end{array} $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.