/*! 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 33 You are watching an object that ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are watching an object that is moving in SHM. When the object is displaced \(0.600 \mathrm{~m}\) to the right of its equilibrium position, it has a velocity of \(2.20 \mathrm{~m} / \mathrm{s}\) to the right and an acceleration of \(8.40 \mathrm{~m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Short Answer

Expert verified
The object will move an additional 0.2887 m from the given point before it momentarily stops and starts to move back to the left.

Step by step solution

01

Understanding SHM

In SHM, we use the formula \(a=-\omega^{2}x\), where \(a\) is acceleration, \(-\omega^{2}\) is the angular frequency squared, and \(x\) is the displacement. Acceleration is directed towards the equilibrium, so it is negative when the displacement, \(x\), is positive. From the problem, we're given \(a=-8.40 \mathrm{~m} / \mathrm{s}^{2}\), and \(x=0.600 \mathrm{~m}\). Therefore, we can solve for \(\omega^{2}\) using the formula: \([-8.40 \mathrm{~m} / \mathrm{s}^{2} / 0.600 \mathrm{~m} = \omega^{2}\). This gives us \(\omega^{2}\approx -14 \mathrm{s}^{-2}\). The angular frequency is always a positive number, so simply take the absolute value: \(\omega=\sqrt{14} \mathrm{s}^{-1}\).
02

Calculating the Amplitude

To find at what displacement position the object will stop momentarily, we need to find the maximum displacement or the amplitude (A) of the oscillation, where the object’s velocity is 0. We can use the following formula derived from the energy conservation law of SHM: \(v=\omega \sqrt{A^{2}-x^{2}}\). Given are \(v=2.20 \mathrm{~m} / \mathrm{s}\), \(x=0.600 \mathrm{~m}\) and \(\omega \approx \sqrt{14} \mathrm{s}^{-1}\). Solving for \(A\), we have \(A=\sqrt{[(2.20 \mathrm{~m} / \mathrm{s}) / \sqrt{14} \mathrm{s}^{-1}]^2 + (0.600 \mathrm{~m})^{2}}\). After calculating, we find that \(A \approx 0.8887 \mathrm{m}\).
03

Finding the additional displacement

The object has already moved 0.600 m to the right. We calculate the additional displacement from this point by subtracting this current displacement from the amplitude. So, it is \( \Delta x = A - x = 0.8887 \mathrm{m} - 0.600 \mathrm{m} = 0.2887 \mathrm{m}\). The object will move an additional 0.2887 m before it momentarily stops and heads back towards equilibrium.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Frequency
In Simple Harmonic Motion (SHM), angular frequency is a key concept that helps determine how an object oscillates over time. It is represented by the Greek letter omega (\( \omega \)). Angular frequency (\( \omega \)) is related to the frequency and period of oscillation, and it's expressed in radians per second. The equation used is \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the object.

The angular frequency also derives from the relationship in the formula \( a = -\omega^2 x \), where \( a \) is the acceleration and \( x \) represents displacement. This shows how acceleration is directly proportional to both the displacement and the square of the angular frequency. Because SHM involves periodic motion, the value of angular frequency helps us understand the rate at which the oscillation occurs.
Displacement
Displacement in SHM refers to how far the object is from its equilibrium position at any given moment. SHM assumes that the displacement is symmetric around the equilibrium point. Displacement can be positive or negative, depending on the direction from the equilibrium.

In mathematical terms, displacement in SHM is often represented as \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is angular frequency, and \( \phi \) is the phase angle. This formula describes how the position of the object varies with time, showing that displacement is a function of both time and amplitude, reaching a maximum at the endpoints of its path.
Amplitude
Amplitude is a measure of the maximum extent of the motion from the equilibrium position in SHM. It's the greatest distance that the object can reach from the center point during its motion. This is symbolized by \( A \) in the equation \( x(t) = A \cos(\omega t + \phi) \).

In the context of the given problem, when the object is at an amplitude, its velocity is zero, meaning it momentarily stops before reversing direction. Amplitude also signifies the total energy in the system because energy in SHM is proportional to the square of the amplitude. Larger amplitudes equate to more energy in the system because the object travels a longer distance during each swing.
Energy Conservation in SHM
In Simple Harmonic Motion, the principle of energy conservation plays a significant role. When an object moves in SHM, its energy switches between kinetic and potential forms. Total mechanical energy remains constant if there’s no external force.
  • Kinetic Energy (KE) is highest at the equilibrium position because velocity is maximum and potential energy is zero.
  • Potential Energy (PE) is highest at the amplitude, where velocity and KE are zero.
The formula for total energy in SHM is \( E = \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 \). Using energy conservation equations helps to solve for unknowns like amplitude and velocity at different displacement points.
Velocity in SHM
Velocity in SHM describes how quickly and in what direction an object moves. Like displacement, it varies with time and oscillates back and forth. In SHM, the instantaneous velocity, \( v(t) \), is the derivative of displacement with respect to time.

The velocity function is given by \( v(t) = -A \omega \sin(\omega t + \phi) \). At the equilibrium position, the velocity is at its peak because the object moves fastest as it passes through the central point. Conversely, at points of maximum displacement (amplitude), velocity is zero since the object changes direction.

By analyzing the velocity in SHM, one can determine both the object's speed and the direction of its motion at any point in its path. Understanding velocity is crucial for predicting future motion in SHM and solving related physics problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{A} 50.0 \mathrm{~g}\) hard-boiled egg moves on the end of a spring with force constant \(k=25.0 \mathrm{~N} / \mathrm{m} .\) Its initial displacement is \(0.300 \mathrm{~m} . \mathrm{A}\) damping force \(F_{x}=-b v_{x}\) acts on the egg, and the amplitude of the motion decreases to \(0.100 \mathrm{~m}\) in \(5.00 \mathrm{~s}\). Calculate the magnitude of the damping constant \(b\).

The jerk is defined to be the time rate of change of the acceleration. (a) If the velocity of an object undergoing SHM is given by \(v_{x}=-\omega A \sin (\omega t),\) what is the equation for the \(x\) -component of the jerk as a function of time? (b) What is the value of \(x\) for the object when the \(x\) -component of the jerk has its largest positive value? (c) What is \(x\) when the \(x\) -component of the jerk is most negative? (d) When it is zero? (e) If \(v_{x}\) equals \(-0.040 \mathrm{~s}^{2}\) times the \(x\) -component of the jerk for all \(t,\) what is the period of the motion?

A \(1.50 \mathrm{~kg}\) mass on a spring has displacement as a function of time given by $$ x(t)=(7.40 \mathrm{~cm}) \cos [(4.16 \mathrm{rad} / \mathrm{s}) t-2.42] $$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{~s} ;\) (f) the force on the mass at that time.

A block with mass \(M\) rests on a friction less surface and is connected to a horizontal spring of force constant \(k .\) The other end of the spring is attached to a wall (Fig. \(\mathbf{P 1 4 . 6 6 )} .\) A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{\mathrm{s}}\). Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

An object is undergoing SHM with period \(0.300 \mathrm{~s}\) and amplitude \(6.00 \mathrm{~cm} .\) At \(t=0\) the object is instantaneously at rest at \(x=6.00 \mathrm{~cm}\). Calculate the time it takes the object to go from \(x=6.00 \mathrm{~cm}\) to \(x=-1.50 \mathrm{~cm}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.