The motion of a particle connected to a spring is described by x= 10sin(πt + π/3). At what time (in s) is the potential energy equal to the kinetic energy?

Answer 1

(1/2) k x^2 = (1/2) m v^2

x= 10sin(πt + π/3)
dx/dt = v = 10π cos(πt + π/3)

x^2 = 100 sin^2(πt + π/3)
v^2 = 100 π^2 cos^2(πt + π/3)

100 k sin^2(πt + π/3) = 100 m π^2 cos^2(πt + π/3)

tan^2 (πt + π/3) = (m/k) π^2

but w^2 = k/m
(2 π/T)^2 = k/m
T^2 = (m/k)(2 π)^2
in our case T occurs when πt = 2 π
or T = 2
so
4 = (m/k)(2 π)^2
or
m/k = 1/π^2

so back to
tan^2 (πt + π/3) = (m/k) π^2 = 1 now
so
πt + π/3 = π/4 (45 degrees)
t = 1/4 - 1/3 no good, negative
so use
πt + π/3 = 3π/4 (second quadrant, tan is -1)
πt + π/3 = 3π/4
t = 3/4 - 1/3 = (9-4)/12 = 5/12
try also 5π/4 etc