Physics

Find the instantaneous velocity of a pendulum at the instant when its bob is at the height equal to half of its maximum height h=(h_max/2) above the equilibrium point. Assume the max velocity of the bob during each oscillation is v_max= 2m/s.

Wouldn't PE=KE
mgh=1/2mv^2
gh=1/2v^2
9.8m/s^2(h_max/2)=1/2v^2
This is as far as I got since I couldn't find v.

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asked by Andrea
  1. The maximum speed is at the bottom. Call the potential energy 0 at the bottom.
    Ke at bottom = (1/2) m v^2 = .5 m (2)^2 = 2 m
    That is also the total energy since PE = 0 at the bottom.
    NOW
    when we go halfway up, the PE change is half the change to the top.
    HOWEVER
    the KE at the top is 0 since it stops to come back.
    THEREFORE
    we have half the total energy as PE and half as KE when we are halfway up
    so
    (1/2) m v^2 = (1/2)(2m)
    .5 v^2 = 1
    v^2 = 2
    v = sqrt 2 = 1.41 m/s

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    posted by Damon

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