/*! 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 17 An oscillator consists of a bloc... [FREE SOLUTION] | 91Ó°ÊÓ

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An oscillator consists of a block attached to a spring ( \(k=\) \(400 \mathrm{~N} / \mathrm{m}) .\) At some time \(t\), the position (measured from the system's equilibrium location), velocity, and acceleration of the block are \(x=0.100 \mathrm{~m}, v=-13.6 \mathrm{~m} / \mathrm{s}\), and \(a=-123 \mathrm{~m} / \mathrm{s}^{2} .\) Calculate (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.

Short Answer

Expert verified
(a) 5.58 Hz, (b) 0.325 kg, (c) 0.102 m.

Step by step solution

01

Find the Angular Frequency

The angular frequency \( \omega \) for a spring-block system is given by the formula \( \omega = \sqrt{\frac{k}{m}} \). First, we need to express the acceleration in terms of angular frequency and displacement: \( a = -\omega^2 x \). Plug in the given values: \( a = -123 \, \mathrm{m/s^2}, \ x = 0.100 \, \mathrm{m} \). Solving for \( \omega \) results in:\[ \omega^2 = \frac{123}{0.100} = 1230 \]So, \( \omega = \sqrt{1230} \).
02

Calculate the Frequency

Now, calculate the frequency \( f \) using the formula \( f = \frac{\omega}{2\pi} \). If \( \omega \approx \sqrt{1230} \approx 35.07 \, \mathrm{rad/s} \), then:\[ f = \frac{35.07}{2\pi} \approx 5.58 \, \mathrm{Hz} \]
03

Determine the Mass of the Block

Use the formula for angular frequency \( \omega = \sqrt{\frac{k}{m}} \) to find the mass \( m \) of the block. Given that \( \omega = 35.07 \, \mathrm{rad/s} \) and \( k = 400 \, \mathrm{N/m} \):\[ 35.07 = \sqrt{\frac{400}{m}} \]Square both sides and solve for \( m \):\[ m = \frac{400}{1230} \approx 0.325 \mathrm{\ kg} \]
04

Calculate the Amplitude of Motion

Use the equations of motion and the given values of acceleration and velocity. The maximum velocity \( v_m = \omega A \) and acceleration \( a = -\omega^2 x \). Solve for amplitude \( A \) using maximum velocity:\[ v = \omega \sqrt{A^2 - x^2} \]Rearranging for \( A \) gives:\[ A = \sqrt{\left(\frac{v}{\omega}\right)^2 + x^2} \]Plugging in \( v = -13.6 \, \mathrm{m/s}, \ \omega \approx 35.07 \, \mathrm{rad/s}, \ x = 0.100 \, \mathrm{m} \):\[ A = \sqrt{\left(\frac{-13.6}{35.07}\right)^2 + 0.100^2} \approx 0.102 \mathrm{\ m} \]

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Key Concepts

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

Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of oscillatory motion where the restoring force acting on an object is directly proportional to its displacement from the equilibrium position, and always acts towards it. One can visualize it like the smooth back and forth motion of a swing.

In SHM, this motion repeats itself in a regular cycle, and it's defined by parameters like amplitude, frequency, and angular frequency. Such motion is periodic and can be described using sine or cosine functions, which are mathematical functions that depend on time.

Understanding SHM is crucial because it provides a foundation for analyzing different physical systems, from simple pendulums to complex mechanical and electronic systems.
Angular Frequency
Angular frequency, represented by the symbol \( \omega \), measures how fast an object is oscillating in Simple Harmonic Motion. It is expressed in radians per second (rad/s). For a mass-spring system, it is determined by the formula:\[\omega = \sqrt{\frac{k}{m}} \]where \( k \) is the spring constant and \( m \) is the mass of the object attached to the spring.

Angular frequency is a crucial concept in understanding oscillations because it relates the physical properties of the mass-spring system to its motion characteristics. A higher angular frequency means the system oscillates faster, while a lower one indicates slower oscillations.

In our calculation, knowing that \( \omega = \sqrt{1230} \approx 35.07 \, \text{rad/s} \) helps us find other properties of the motion, such as frequency and amplitude.
Mass-Spring System
The Mass-Spring System is a classic example of SHM, which consists of a block (mass \( m \)) attached to a spring with a spring constant \( k \).

In ideal conditions, when the block is displaced and released, it oscillates about the equilibrium position under the influence of the restoring force provided by the spring. The spring force acts in the opposite direction to the displacement and follows Hooke’s Law:
  • \( F = -kx \)
The negative sign indicates that the force is always trying to bring the block back to equilibrium.

By understanding a mass-spring system, one can derive essential formulas to calculate the system's properties, including angular frequency \( \omega \), frequency \( f \), and amplitude \( A \). Recognizing that \( m = 0.325 \text{ kg} \) shows how the physical properties influence the system's motion.
Amplitude Calculation
Amplitude \( A \) in Simple Harmonic Motion refers to the maximum displacement from the equilibrium position. It tells us the "size" of the oscillation. In mathematical terms, amplitude is a crucial value that affects how we describe the system's motion.

To calculate amplitude when the maximum velocity and position are known, we use the relation:\[A = \sqrt{\left(\frac{v}{\omega}\right)^2 + x^2}\]Here, \( v \) is the velocity, \( \omega \) is the angular frequency, and \( x \) is the position from equilibrium. Substituting the values for this system provides \( A \approx 0.102 \, \text{m} \).
  • This value describes the furthest distance the mass moves from its rest position.
Understanding amplitude is essential for identifying motion characteristics in any oscillating system and is a key part of describing the system's dynamics accurately.

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Most popular questions from this chapter

Suppose that a simple pendulum consists of a small \(60.0 \mathrm{~g}\) bob at the end of a cord of negligible mass. If the angle \(\theta\) between the cord and the vertical is given by $$ \theta=(0.0800 \mathrm{rad}) \cos [(4.43 \mathrm{rad} / \mathrm{s}) t+\phi] \text { , } $$ what are (a) the pendulum's length and (b) its maximum kinetic energy?

A block weighing \(10.0 \mathrm{~N}\) is attached to the lower end of a vertical spring \((k=200.0 \mathrm{~N} / \mathrm{m})\), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of \(2.00 \mathrm{~J}\) as it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates?

An oscillating block-spring system takes \(0.75 \mathrm{~s}\) to begin re- peating its motion. Find (a) the period, (b) the frequency in hertz, and (c) the angular frequency in radians per second.

A \(3.0 \mathrm{~kg}\) particle is in simple harmonic motion in one dimension and moves according to the equation $$ x=(5.0 \mathrm{~m}) \cos [(\pi / 3 \mathrm{rad} / \mathrm{s}) t-\pi / 4 \mathrm{rad}] $$ with \(t\) in seconds. (a) At what value of \(x\) is the potential energy of the particle equal to half the total energy? (b) How long does the particle take to move to this position \(x\) from the equilibrium position?

A grandfather clock has a pendulum that consists of a thin brass disk of radius \(r=15.00 \mathrm{~cm}\) and mass \(1.000\) kg that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in Fig. \(15.56 .\) If the pendulum is to have a period of \(2.000 \mathrm{~s}\) for small oscillations at a place where \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\), what must be the rod length \(L\) to the nearest tenth of a millimeter?

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