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## Homework Statement

A mass is oscillating on a spring with simple harmonic motion. Its amplitude is 0.80m and its maximum speed is 1.5m/s, at the point of equilibrium. What is the speed of the mass at 0.60m?

Is there enough information to answer this question? The answer given is 0.75m/s, but I can't figure out how to get there.

## Homework Equations

Ep = kx

^{2}/2

Ec = mv

^{2}/2

Em = Ep + Ec

## The Attempt at a Solution

I've made many attempts at this, all completely wrong. I always end up trying to factorize (my teacher told us that the answer involves factorizing) but it goes absolutely haywire.

Em = Em'

Ep

_{max}= Ec

_{max}

kx

_{max}

^{2}/2 = mv

_{max}

^{2}/2 (the 1/2 cancels for each formula)

v

_{max}/x

_{max}= √k/m (which should be constant)

(1,5m/s)

^{2}/(0.80m)

^{2}= √k/m

1.875(not sure what unit to put here) = √k/m ∴ k/m = 3.516

kx

_{2}+ mv

^{2}= mv

_{max}

^{2}(again the 1/2 cancels)

mv

_{max}

^{2}- mv

^{2}= kx

^{2}

m(v

_{max}

^{2}- v

^{2}) = kx

^{2}

v

_{max}

^{2}- v

^{2}= kx

^{2}/m

k/m = (v

_{max}

^{2}- v

^{2})/x

^{2}

If the k/m stays constant, which I believe it should, it should be an easy substitution to find the v I'm looking for.

After manipulating the formula I got, I'm left with

v = √-(k/m - v

_{max}

^{2}/x

^{2})(x

^{2})

After substituting, I get 0.99m/s.