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## Homework Statement

(Forgive me if I use certain terms wrong; I'm learning all of this in French)

A branch at the top of a tree is swinging back and forth with a simple harmonic motion. Its amplitude is 0.80m. Its maximum speed at the point of equilibrium is 1.5m/s. What is the speed of the branch at 0.60m?

The answer according to the back of the book is 0.75m/s but I need to know how to get there.

## Homework Equations

Em = Em'

Ep + Ec = Ep' + Ec'

Ep = kx

^{2}/2

Ec = mv

^{2}/2

## 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. It would take too long to type out even one attempt.

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.

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