# Energy Problem involving simple harmonic motion

1. Dec 11, 2007

### myoplex11

1. The problem statement, all variables and given/known data
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 ?

2. Relevant equations

KE = 1/2 mv^2 PE=1/2 Kx^2
3. The attempt at a solution

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 iam i going wrong.

2. Dec 11, 2007

### Kurdt

Staff Emeritus
There appear to be a lot of algebraic errors throughout. Where has the m and k gone from the equation toward the end?

EDIt: up to the this point you seem to be fine as far as I can tell:

m/k * pie^2 = tan^2(piet +pie/2 )

I just want to know how you got rid of the m and k in the next step?

3. Dec 11, 2007

### myoplex11

w(angular frequency) = (k/m)^0.5 so
1/w=m/k
siince w = pie
m/k = 1/pie

4. Dec 11, 2007

### Kurdt

Staff Emeritus
I tried a different approach and got the same answer. since the motion is SHM one would think that value +$n2\pi$ should work.

5. Dec 11, 2007

### myoplex11

what is n2PI is it equal to m/k ?

6. Dec 11, 2007

### rl.bhat

You can wright PE = 1/2*m*w^2*x^2 and KE = 1/2*m*w^2*(A^2 - x^2)
When PE = KE, we get 2x^2 = A^2 or x = A/sqrt2. Put this value in the equation of SHM.
A/sqrt2 = Asin(pi*t + Pi/2 ) or 1/sqrt2 = sin(pi*t + pi/2). Sin(pi/4) = sin(3*pi/4) = 1/sqrt2 To avoide negative time, take sin(3pi/4) = sin(pi*t + pi/2)
That gives you t = 1/4 s.

Last edited: Dec 11, 2007
7. Dec 11, 2007

### myoplex11

the anwser is 0.9 s