Simple Harmonic Motion + Regular Kinematics

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SUMMARY

The discussion focuses on a particle's motion along the x-axis, initially positioned at 0.250 m with a velocity of 0.200 m/s and an acceleration of -0.290 m/s². After 5.80 seconds of constant acceleration, the particle's position is calculated to be -3.4678 m, and its velocity is -1.482 m/s. The subsequent motion involves simple harmonic motion (SHM) for another 5.80 seconds, where the participant attempts to apply the formula X(t) = A cos(ωt) with an amplitude of 0.250 m and angular velocity derived from the initial conditions. However, the participant struggles to incorporate the effects of constant acceleration into the SHM equations.

PREREQUISITES
  • Understanding of kinematics, specifically equations of motion under constant acceleration.
  • Knowledge of simple harmonic motion (SHM) equations, including X(t) = A cos(ωt) and v(t) = ωA sin(t).
  • Familiarity with angular velocity calculations and its application in SHM.
  • Ability to analyze motion in one dimension along the x-axis.
NEXT STEPS
  • Study the derivation and application of simple harmonic motion equations in varying conditions.
  • Learn how to combine constant acceleration with harmonic motion in practical scenarios.
  • Explore the relationship between amplitude, angular frequency, and energy in SHM.
  • Investigate the effects of initial conditions on the motion of a particle in SHM.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and simple harmonic motion, as well as educators seeking to clarify these concepts for their students.

fnaarfnaar
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Homework Statement


A particle moves along the x axis. It is initially at the position 0.250 m,
moving with velocity 0.200 m/s and
acceleration -0.290 m/s2.
Suppose it moves with constant acceleration for 5.80 s.

Position of Particle after 5.80 Seconds - -3.4678 m.
Velocity of Particle After 5.80 Seconds -1.482 m/s

Next, assume it moves with simple harmonic motion for 5.80 s and x = 0 is its equilibrium position.

What is its position at the end of this time interval?
What is its velocity at the end of this time interval?

Homework Equations

Attempts :

I attempted to use the formula X(t)= A cos (wt), with A being the amplitude 0.250 (because a spring can't stretch farther than its intial position) and w = angular velocity which I used 0.2 / .25.

For the velocity I used the formula- v(t)=w A sin (t)

From this point I was completely lost because I could not find a way to incorporate the constant acceleration. Simple Harmonic Motion equations

if you could leave an explanation, it's be greatly appreciated!
 
Last edited:
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fnaarfnaar said:

Homework Statement


A particle moves along the x axis. It is initially at the position 0.250 m,
moving with velocity 0.200 m/s and
acceleration -0.290 m/s2.
Suppose it moves with constant acceleration for 5.80 s.

Position of Particle after 5.80 Seconds - -3.4678 m.
Velocity of Particle After 5.80 Seconds -1.482 m/s

Next, assume it moves with simple harmonic motion for 5.80 s and x = 0 is its equilibrium position.

What is its position at the end of this time interval?
What is its velocity at the end of this time interval?


Homework Equations



Simple Harmonic Motion equations

if you could leave an explanation, it's be greatly appreciated!
You need to make an attempt at solving the problem yourself and showing us what you did before you can receive assistant, per the forum rules.
 

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