Simple Harmonic Motion + Regular Kinematics

AI Thread Summary
The discussion centers on a particle's motion along the x-axis, initially at 0.250 m with a velocity of 0.200 m/s and an acceleration of -0.290 m/s² for 5.80 seconds, resulting in a position of -3.4678 m and a velocity of -1.482 m/s. The next phase involves the particle transitioning to simple harmonic motion (SHM) for another 5.80 seconds, with the equilibrium position set at x = 0. The user attempted to apply SHM formulas, using amplitude and angular velocity, but struggled to incorporate the effects of the previous constant acceleration. Clarification on how to combine these concepts in calculations is requested, adhering to forum rules that require an initial attempt at problem-solving.
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!
 
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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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