by The Merf
Tags: oscillation, spring, suspended mass
 P: 10 1. The problem statement, all variables and given/known data A 0.27 kg mass is suspended on a spring that stretches a distance of 4.9 cm. The mass is then pulled down an additional distance of 12.5 cm and released. What's the displacement from the equilibrium position with the mass attached (in cm) after 0.42 s? Take up to be positive and use g = 9.81 m/s^2 m=.27kg t=.42s g=9.81m/s^2 I don't know how to represent it, but I believe that 4.9 is the equilibrium. 12.5cm is the initial displacement (maybe x0?) 2. Relevant equations I have no idea, this is what I need help with. maybe u=1/2kx^2, but I don't have a k, and I don't need u, but it is the only spring equation I know 3. The attempt at a solution
 P: 10 I also found the fallowing equations dealing with harmonic oscillation: $$x(t)=A\sin(\omega t)+B\cos(\omega t)$$ $$\omega^2 =\frac{k}{m}$$ i'm not sure what the w is
 P: 10 ooh, and this one look very promissing: [ctex] x(t)=\frac{v_0}{\omega}\sin(\omega t)+x_0\cos(\omega t)\;. [/ctex] I know that v0 is 0 so that wipes out the first half, I think x0 is 12.5, but where does the 4.9 come in and what is the w? hmm, the image doesn't seem to work so i'm going to try writing it out: x(t)=((v0)/w)sin(wt) + (x0)cos(wt)
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