A 1-kg block executes simple harmonic motion with an amplitude A = 15 cm. In 6.8 sec, the block completes 5-oscillations. Determine the kinetic energy of the oscillator, K = ?, at a position where the potential energy is twice the kinetic energy (U = 2K).
period T = 6.8 / 5 = 1.36 seconds
so
x = 0.15 sin (2 pi t/1.36) = 0.15 sin 4.62 t
v = dx/dt = 0.693 cos 4.62 t
Ke = (1/2) m v^2 = (1/2) m .480 cos^2 4.62 t
thus maximum Ke = 0.240 m = total sum of Ke and U = constant
so
0.240 m = 0.240 m (cos^2 4.62 t + sin^2 4.62t) because cos^2 + sin^2= 1
so
U = 0.240 m sin 4.62 t
when is U = 2 Ke ?
sin 4.62 t = 2 cos 4.62 t
tan 4.62 t = 2
4.62 t = 63.4 degrees * pi/180 = 1.11 radians
Ke = (1/2) m v^2 = (1/2) m .480 cos^2 4.62 t = 0.240(1) cos^2 63.4 deg
= 0.107