Question

The motion of a particle is described by x = 10 sin (piet +pie/2 ). At what time ( in second ) is the potential energy equal to the kinetic energy ?
here is what i have done so far where am i going wrong
V=pie10cos(piet + pie/3)
V^2 = pie^2 100cos^2(piet+ pie/3)
KE = mpie5ocos^2 (piet + pie/3)
X^2 =100sin^2(piet +pie/2 )
PE =k50 sin^2(piet +pie/2 )
KE=PE
pie^2 100cos^2(piet+ pie/3) =K50 sin^2(piet +pie/2 )
m/k * pie^2 = tan^2(piet +pie/2 )
take square root of both sides of equation
1/pie * pie = tan(piet + pie/3)
piet + pie/3 =tan-1(1)
solving for t i get a negative value where i am i going wrong

Answer 1

You said pi/2 at first, then changed to pi/3
I assume that we are talking about a spring here so
PE = (1/2)kx^2
then
v = 10 pi cos (pi t + pi/2)
v^2 = 100 pi^2 cos^2 (pi t + pi/2)
KE = (1/2) m v^2 = 50 m pi^2 cos^2(pi t +pi/2)
so sort of agree with you
Now PE = (1/2) k x^2 =
(1/2)k 100 sin^2( pi t + pi/2)
so when does PE = KE?
50 m pi^2 cos^2 pi t + pi/2) =50 k sin^2(pi t + pi/2)

m pi^2 cos^2 = k sin^2
tan^2 (pi t + pi/2 ) =pi^2 m/k
tan (pi t +pi/2) = +/- pi sqrt(m/k)
two solutions:
t =-1/2 + (1/pi)tan^-1(pi sqrt(m/k))
t =-1/2 - (1/pi)tan^-1(pi sqrt(m/k))

Now sqrt (m/k) means something
wo the natural frequency = sqrt (k/m)
= 2 pi fo = 2 pi/T where T is the period of free vibration
so
sqrt (m/k) = T/ 2 pi
so use that