Question on Simple Harmonic Motion.

[Wesc]

Homework Statement

A 100g particle hangs freely at rest on the end of a spring of stiffness 10N/m. If
the particle is projected upwards with a speed of 2m/s, find the time taken until
it first comes to rest and the distance travelled.

Homework Equations

Well, there's F = -k.x and of course the classic F = ma .. and the normal linear motion equations? And the conservation of energy too I think...

The Attempt at a Solution

Ok, well first off this is probably wrong because I suck at Simple Harmonic Motion, but this is what I did:

I started off by finding the distance between L0 and the equilibrium position, by using m.a = k.x ... and got x = (0.1)(9.8)/(10) =0.098.

Hence, I did the following:
(0.5)(k)(x^2) + (0.5)(m)(u^2) = (0.5)(m)(V^2) ... and got V to equal 2.2272 m/s.

Then, I assumed that normal motion would resume and used "v=u+at" and "(v^2-u^2)/2a = s" to calculate the rest. I obtained t = 0.227 s and s = 0.2531 m
There are no answers available for this question unfortunately, and I've an exam on this topic tomorrow, so I was hoping that someone on this that understands SHM could help me if I'm doing this wrong? Thank you.

 

[vela]

Homework Statement

A 100g particle hangs freely at rest on the end of a spring of stiffness 10N/m. If
the particle is projected upwards with a speed of 2m/s, find the time taken until
it first comes to rest and the distance travelled.

Homework Equations

Well, there's F = -k.x and of course the classic F = ma .. and the normal linear motion equations? And the conservation of energy too I think...

The Attempt at a Solution

Ok, well first off this is probably wrong because I suck at Simple Harmonic Motion, but this is what I did:

I started off by finding the distance between L0 and the equilibrium position, by using m.a = k.x ... and got x = (0.1)(9.8)/(10) =0.098.

Hence, I did the following:
(0.5)(k)(x^2) + (0.5)(m)(u^2) = (0.5)(m)(V^2) ... and got V to equal 2.2272 m/s.

You didn't need to do any of this. The first sentence says the mass "hangs freely at rest." This means that the spring has already been stretched to the new equilibrium position, and the mass is just sitting there.

Then, I assumed that normal motion would resume and used "v=u+at" and "(v^2-u^2)/2a = s" to calculate the rest. I obtained t = 0.227 s and s = 0.2531 m

These equations only apply to cases of constant acceleration. They don't apply to simple harmonic motion because the force exerted by the spring is constantly changing.


Take a look at the top diagram on http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html. Based on the givens in this problem, where in the cycle is the particle initially? Is it at the top, the bottom, the middle, or somewhere else? Explain your reasoning for your answer.

 

[haruspex]

Hence, I did the following:
(0.5)(k)(x^2) + (0.5)(m)(u^2) = (0.5)(m)(V^2) ... and got V to equal 2.2272 m/s.

But what do you think this speed you have calculated represents?