posted by Zoe .
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?
v2 +ω^2x^2 =ω^2A^2
vmax =ω A and v = ωA/2?
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