Posted by **shan** on Wednesday, December 12, 2007 at 2:14am.

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

- physics -
**Damon**, Wednesday, December 12, 2007 at 9:48am
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

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