(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant 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!!

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# Homework Help: Simple Harmonic Motion + Regular Kinematics

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