- #1

SplinterIon

- 17

- 0

[tex] \frac {d^2x} {dt^2} = -Qx + R [/tex]

I'm supposed to find a solution expressing x in terms of t, given that position is (5,0) at t=0 and speed=0. The next time the particle is at (5,0) it has traveled a distance of 12 units and t = Pi. I'm supposed to solve for Q and R as well and have x and t as the only 'variables'.

The differential equation isn't that hard to work out - solving the characteristic eqn - leading to complex roots - I get

[tex] x = A \sin{(\sqrt{Q}t)} + B \cos{(\sqrt{Q}t)}\ +\frac{R}{Q} [/tex]

I believe I'm correct in saying A = 0 since taking [tex]\frac{dx}{dt}[/tex] and setting it equal to 0 with t=0 gives [tex]\sqrt(Q)*A = 0 [/tex]. From this point on I'm a little stuck - as nothing I do seems to be able to solve for B or R and Q - I don't even know how to bring in the 12 to help out. A little help would be appreciated.