A particle executes simple harmonic motion with an amplitude of 9.00 cm. At what positions does its speed equal two thirds of its maximum speed?

I got:

v2 +ω^2x^2 =ω^2A^2

vmax =ω A and v = ωA/2?

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  1. Since it executes SHM, the sum of potential energy and kinetic energy is a constant, with PE proportional to x^2 and KE proportional to V^2. This means

    (x/A)^2 + (V/Vmax)^2 = 1

    where A is the amplitude and x is the distance from the mean position.

    When V/Vmax = 2/3,
    (x/A)^2 = 1 - 4/9 = 5/9
    x/A = (sqrt5)/3 = (+or-) 0.7454

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