Question abou Simple Harmonic Motion

jcsolis
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Homework Statement



A spring mass system has the following parameters: Mass m=1 kg, spring constant k=100 N/m. The oscillations are started by initially displacing the mass so that the spring is compressed by 0.5 cm and releasing it from rest. What is the speed of the mass (in m/s) when it crosses the equilibrium position?



Homework Equations



x(t)=A sin(wt+phi)
v(t)=Aw^2 cos(wt+phi)

The Attempt at a Solution



Ok, so the spring is compressed by 0.5 cm from its equilibrium position. So when x(t)=0.5 cm, v(t)=0 at t=0

after solving the equation for motion x(t) I got that A=0.5 cm

I don´t know what is next, can somebody help me?
 
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jcsolis said:
v(t)=Aw^2 cos(wt+phi)

I think that your omega shouldn't be squared here but you should check it also remember that as the spring passes through its equilibrium position it would have maximum speed and this can only happen when cos(wt+phi) = 1. (well + or - 1)

Also remember that there's an equation which links K to omega. I'm not sure if you'd have to derive this formula the question doesn't ask you to but it probably depends on how many marks its worth, i'd guess you can just use it though.
 
Last edited:
I'm not sure whether the question gave you those equations or not, but if it didn't then Sanitykey is right; the omega shouldn't be squared, there should be a negative sign (to keep direction), and the equation of x(t) should correspond to cosine and v(t) corresponds to sine.
If you apply a certain amount of force to a system (do work to it), then whatever the initial displacement you gave it is going to be the amplitude. So if you pulled the mass .5 cm from the equilibrium point, then the spring isn't going to displace further than + or - .5cm (a bound system).
 

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