/*! 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 65 An object is undergoing SHM with... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An object is undergoing SHM with period 1.200 s and amplitude 0.600 \(\mathrm{m}\) . At \(t=0\) the object is at \(x=0\) and is moving in the negative \(x\) -direction. How far is the object from the equilibrium position when \(t=0.480 \mathrm{s} ?\)

Short Answer

Expert verified
0.359 meters.

Step by step solution

01

Determine the Equation for SHM

The equation of motion for Simple Harmonic Motion (SHM) is given by the formula: \[ x(t) = A \cdot \cos(\omega t + \phi) \]where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. In this problem, we are given that at \( t = 0 \), the object is at \( x = 0 \) moving in the negative direction. This suggests that the cosine function is at a point where it equals zero, and the derivative is negative. This scenario corresponds to \( \phi = \frac{\pi}{2} \).
02

Calculate the Angular Frequency

The angular frequency \( \omega \) is related to the period \( T \) by the formula: \[ \omega = \frac{2\pi}{T} \]Given that the period \( T = 1.200 \) seconds, plug in the values:\[ \omega = \frac{2\pi}{1.200} \approx 5.236 \text{ radians/second} \]
03

Substitute Known Values into the SHM Equation

Substitute the amplitude, angular frequency, and phase constant into the equation:\[ x(t) = 0.600 \cdot \cos(5.236t + \frac{\pi}{2}) \]This is the SHM equation specific to this problem.
04

Calculate Position at t = 0.480 s

Using the SHM equation, calculate the position at \( t = 0.480 \) seconds:\[ x(0.480) = 0.600 \cdot \cos(5.236 \times 0.480 + \frac{\pi}{2}) \]Using a calculator for the cosine function:\[ x(0.480) = 0.600 \cdot \cos(2.513 + \frac{\pi}{2}) \]\[ \cos(2.513 + \frac{\pi}{2}) = \cos(4.084) \approx -0.598 \]Then,\[ x(0.480) = 0.600 \times -0.598 \approx -0.359 \text{ meters} \]Since we're considering the distance from equilibrium, the absolute value is:\[ \text{Distance} = 0.359 \text{ meters} \]

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.

Understanding Angular Frequency
Angular frequency is a crucial element of Simple Harmonic Motion (SHM), which describes how quickly an object moves through its cycle. It is denoted by the symbol \( \omega \). This concept is similar to the frequency we use in everyday terms, but with a twist—it measures the rate of rotation in radians per second rather than cycles per second or hertz.

The formula for angular frequency \( \omega \) is:
  • \( \omega = \frac{2\pi}{T} \)
Here, \( T \) represents the period of the motion, which is the time taken to complete one full cycle. By using this formula, you can directly relate the period and angular frequency, providing insight into the speed of the oscillation.

In our problem, the period is given as 1.200 seconds, leading to an angular frequency of approximately 5.236 radians/second. Understanding angular frequency allows you to grasp how quickly the system undergoes each oscillation.
Defining the Period
The period is the duration required for one full cycle of movement in Simple Harmonic Motion. It is an essential concept that highlights the regularity and predictability of oscillations.

The period \( T \) of SHM is often measured in seconds and is inversely related to the frequency, indicating how long it takes for the object to return to a particular point in its oscillatory path. This regular interval underscores the periodic nature of SHM.

In the example, the period is 1.200 seconds, meaning every 1.200 seconds, the object completes one full oscillation. Knowing the period helps us determine not only the object’s speed through its path but also its timing and rhythm in the oscillatory motion.
The Role of Equilibrium Position
In Simple Harmonic Motion, the equilibrium position is the point where the net force acting on the object is zero. It is generally the midpoint of the motion, where the object would remain at rest if not for any disturbance.

In essence, the equilibrium position acts as a balance point in the system, from which the object oscillates back and forth. At this point, the potential energy is minimal, and the kinetic energy is at its peak. Any deviation from this point results in forces that tend to restore the object back to equilibrium.

In our scenario, understanding the equilibrium position helps us calculate how far the object is from this center point at any given time, such as at \( t = 0.480 \) seconds.
Grasping Amplitude
Amplitude is a significant measure in Simple Harmonic Motion, representing the maximum displacement from the equilibrium position. It reflects the extent of the object's oscillation and gives insight into the energy within the system.

Amplitude is denoted by \( A \), and it expresses how far the object moves away from its resting or equilibrium state during its motion. In the given example, the amplitude is 0.600 meters, indicating the farthest distance the object reaches from its equilibrium point.

Remember, the larger the amplitude, the more "intense" the oscillation is, typically corresponding to higher energy levels in the system. Therefore, understanding amplitude is key to analyzing the overall dynamics and behavior of SHM systems.

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

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.120 \(\mathrm{m} .\) The maximum speed of the block is 3.90 \(\mathrm{m} / \mathrm{s}\) . What is the maximum magnitude of the acceleration of the block?

The tip of a tuning fork goes through 440 complete vibrations in 0.500 s. Find the angular frequency and the period of the motion.

A proud deep-sea fisherman hangs a 65.0 -kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) . (a) Find the force constant of the spring. The fish is now pulled down 5.00 \(\mathrm{cm}\) and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

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

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 \(\mathrm{m}\) and the engine runs at 4500 \(\mathrm{rev} / \mathrm{min}\) , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass \(0.450 \mathrm{kg},\) what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

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.