Solving for Potential & Kinetic Energy: x=10sin(pi*t+pi/3)

[jrodss]

1. The motion of a particle is described by x=10sin(pi*t+pi/3). At what time in s is the potential energy equal to the kinetic energy?

2. I know these equations come into play
Kinetic Energy
K=1/2*m*ω^2*A^2*sin^2(ωt+φ)
Kmax=1/2 k A^2
K=1/2 mv^2

Potential Energy
U=1/2*k*A^2*cos^2(ωt+φ)
U=1/2 kx^2

3. The Attempt at a Solution

Well I know A=10, w=pi, T=pi seconds I am really stuck on where to go from here I think I am trying to make it too complicated. Any pointers would be great. Also if someone could explain what the phase constant does to the equation it would be very helpful.

 

[rock.freak667]

you want where

$$\frac{1}{2}mv^2 = mgx$$


m will cancel out and you know that v= dx/dt.

then remember that cos²A+sin²A=1

 

[jrodss]

from that suggestion I end up with:

1/2*(10pi*cos(pi*t+pi/3))^2=9.8*10*sin(pi*t+pi/3)

Simplifying gets

10pi^2/9.8=sin(pi*t+pi/3)/cos(pi*t+pi/3)^2

I am still jammed up

 

[rock.freak667]

10pi^2/9.8=sin(pi*t+pi/3)/cos(pi*t+pi/3)^2

I am still jammed up

(10pi^2/9.8)cos²(pi*t+pi/3)=sin(pi*t+pi/3)

Now

$$cos^2(\pi t + \frac{\pi}{3}) + sin^2(\pi t + \frac{\pi}{3})=1$$


sub for [itex]cos^2(\pi t + \frac{\pi}{3})[/tex]

then you will have a quadratic in $sin(\pi t + \frac{\pi}{3})$

 

[rl.bhat]

KE = PE
0.5mv^2 = 0.5kx^2
v = dx/dt = Aωsin(ωt + φ)
v^2 = A^2ω^2sin^2(ωt + φ) = ω^2(A^2 - x^2)
So 0.5mv^2 = 0.5kx^2
mω^2(A^2 - x^2) = kx^2 = mω^2(x^2)
or A^2 - x^2 = x^2 or A^2 = 2x^2 or x^2 = A^2/2 or x = A/(2)^1/2
Substitute this value in the given equation and solve for t. You can see how the phase angle helps to find t.
In this problem g has no roll at all.