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Chapter 13: Problem 11

A \(2.00-\mathrm{kg}\) , frictionless block is attached to an ideal spring with force constant 300 \(\mathrm{N} / \mathrm{m}\) . At \(t=0\) the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 \(\mathrm{m} / \mathrm{s}\) . Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

Short Answer

Expert verified
(a) Amplitude: 0.979 m, (b) Phase angle: \(\frac{3\pi}{2} \), (c) Position: \(x(t) = 0.979 \cos(12.247t + \frac{3\pi}{2})\)."

Step by step solution

01

Understand the Problem

We need to find the amplitude, phase angle, and position function for a block attached to a spring with specified initial conditions. This is a classic simple harmonic motion problem.
02

Analyze the Initial Conditions

Given that at time \( t = 0 \), the block is moving in the negative direction with speed \( v(0) = -12.0 \ m/s \), and the spring is neither stretched nor compressed at this point. The mass of the block is \( m = 2.00 \ kg \), and the spring constant is \( k = 300 \ N/m \).
03

Calculate Angular Frequency

The angular frequency \( \omega \) of the motion is given by \( \omega = \sqrt{\frac{k}{m}} \). Substituting the given values: \[\omega = \sqrt{\frac{300}{2.00}} = \sqrt{150} = 12.247 \ rad/s.\]
04

Set Up the Position Function

The position function for simple harmonic motion has the form \[ x(t) = A \cos(\omega t + \phi). \]We will use the initial conditions to find values for \( A \) and \( \phi \).
05

Determine Amplitude

At \( t = 0 \), the displacement is zero, so \( x(0) = A \cos(\phi) = 0 \). This implies \( \phi = \frac{\pi}{2} \) or \( \phi = \frac{3\pi}{2} \). We check this by considering velocity.
06

Use Velocity to Solve for \( \phi \)

The velocity equation is \[ v(t) = -A \omega \sin(\omega t + \phi). \]At \( t=0 \), \( v(0) = -12 m/s = -A \omega \sin(\phi) \). With \( \omega = 12.247 \), \[ -12 = -A \times 12.247 \times \sin(\phi). \]Solving gives \( A \cdot \sin(\phi) = 0.979 \). Since \( \phi = \frac{3\pi}{2} \) satisfies \( \sin(\phi) = -1 \), we confirm \( \phi = \frac{3\pi}{2}. \)
07

Determine Amplitude with \( \phi \)

Having found \( \sin(\phi) = -1 \), we can find \( A \) as follows:\[12 = A \times 12.247 \times 1 \Rightarrow A = \frac{12}{12.247} = 0.979.\]
08

Write Function for Position

The position as a function of time is: \[ x(t) = 0.979 \cos(12.247 t + \frac{3\pi}{2}). \]

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Key Concepts

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

Angular Frequency
Angular frequency is a fundamental concept in simple harmonic motion, often denoted by the Greek letter \( \omega \). It describes how quickly an object oscillates back and forth in a system, measured in radians per second. For a mass-spring system, angular frequency depends on both the spring constant \( k \) and the mass \( m \) of the object.
  • The formula to calculate angular frequency is \( \omega = \sqrt{\frac{k}{m}} \).
  • Given a spring constant \( k = 300 \, \text{N/m} \) and a mass \( m = 2.00 \, \text{kg} \), the calculated angular frequency is \( \omega = \sqrt{\frac{300}{2.00}} = \sqrt{150} = 12.247 \, \text{rad/s} \).
Angular frequency gives insight into the system's responsiveness: a higher \( \omega \) means faster oscillations. Understanding \( \omega \) helps in determining other quantities like the velocity and position of the mass over time.
Amplitude
Amplitude in simple harmonic motion is the maximum displacement from the equilibrium position. It represents the farthest point that the oscillating object reaches in either direction. In this context, amplitude, denoted as \( A \), can be found using initial velocity conditions.
  • The expression for velocity in such motion is \( v(t) = -A \omega \sin(\omega t + \phi) \).
  • At the initial point \( t = 0 \), the velocity \( v(0) = -12 \, \text{m/s} \).
  • With \( \omega = 12.247 \, \text{rad/s} \) and \( \sin(\phi) = -1 \) (at \( \phi = \frac{3\pi}{2} \)), the amplitude \( A \) is calculated as \( A = \frac{12}{12.247} = 0.979 \).
This means the block's maximum displacement from the equilibrium is 0.979 meters.
Phase Angle
Phase angle \( \phi \) in simple harmonic motion provides information on the starting point of the oscillation relative to the equilibrium position. It determines the initial state of motion at \( t = 0 \).
  • The initial condition for position \( x(0) = A \cos(\phi) \) leads to the solution that \( \cos(\phi) = 0 \), indicating \( \phi \) could be \( \frac{\pi}{2} \) or \( \frac{3\pi}{2} \).
  • Given the initial negative velocity of \( -12 \, \text{m/s} \), \( \sin(\phi) \) must be \(-1\) to satisfy the velocity equation \( v(0) = -A \omega \sin(\phi) \).
  • This results in \( \phi = \frac{3\pi}{2} \), indicating that the block starts its motion at its maximum amplitude in the negative direction.
Understanding \( \phi \) is crucial for accurately writing the position function.
Position Function
The position function in simple harmonic motion describes how the position of the block changes over time. It is expressed as \( x(t) = A \cos(\omega t + \phi) \), where \( A \), \( \omega \), and \( \phi \) are the amplitude, angular frequency, and phase angle, respectively.
  • The derived position function for this specific problem is \( x(t) = 0.979 \cos(12.247 t + \frac{3\pi}{2}) \).
  • This function gives the block's position at any time \( t \), taking into account the initial velocity and position conditions.
  • The positive or negative sign of the cosine function will indicate the direction of the block relative to the equilibrium point.
The position function is pivotal for predicting the behavior of the system over time, providing insights into how the block oscillates back and forth.

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Most popular questions from this chapter

A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm}\) . (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding on to the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 \(\mathrm{kg}\) , and a \(175-\mathrm{kg}\) sack of gravel sits on the middle of it. The beam is oscillating in SHM, with an amplitude of 40.0 \(\mathrm{cm}\) and a frequency of 0.600 cycles \(/ \mathrm{s}\). (a) The sack of gravel falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the gravel instead falls off when the beam has its maximum speed, what are the frequency and amplitude of the subsequent SHM of the beam?

A \(0.400-\mathrm{kg}\) object undergoing \(\mathrm{SHM}\) has \(a_{x}=-2.70 \mathrm{m} / \mathrm{s}^{2}\) when \(x=0.300 \mathrm{m}\) . What is the time for one oscillation?

Two pendulums have the same dimensions (length \(L )\) and total mass \((m) .\) Pendulum \(A\) is a very small ball swinging at the end of a uniform massless bar. In pendulum \(B,\) half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

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.)
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