Solving a Particle's Differential Equation: Finding x(t) and Q, R

AI Thread Summary
The discussion revolves around solving the differential equation of motion for a particle, expressed as d²x/dt² = -Qx + R, with initial conditions at t=0 and t=π. The solution involves finding constants Q and R while ensuring the particle's position and speed are correctly accounted for. The user derives the general solution x = A sin(√Qt) + B cos(√Qt) + R/Q and determines that A = 0, leading to further calculations for B and R. The calculations reveal that Q is determined to be 4 and B is found to be 3, resulting in R being 8. The discussion emphasizes the importance of not providing complete solutions but rather guiding the user through the problem-solving process.
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Given a particle whose equation of motion is

\frac {d^2x} {dt^2} = -Qx + R

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

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

I believe I'm correct in saying A = 0 since taking \frac{dx}{dt} and setting it equal to 0 with t=0 gives \sqrt(Q)*A = 0. 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.
 
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Is this actually 2-d? what do you know about y?

At t=0, x=5 = B + R/Q (presuming your soln is ok)
at t=pi, x=5 = B cos(RootQpi) +R/Q
subtract these to solve for Q
find out when v=0 ; then, x=-1 (why?) R=?
 
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At t=0, x=5 = B + R/Q (presuming your soln is ok)
at t=pi, x=5 = B cos(RootQpi) +R/Q

subtracting one from t'other,

0 = B - Bcos(√Q.pi)
B(1 - cos(√Q.pi)) = 0
cos(√Q.pi) = 1
√Q.pi = 0, 2pi, 4p, ...
√Q = 0, 2, 4, ...
Q = 0, 4
Q = 4
====

Then,

x = Bcos(2t) +R/Q
at t = 0, x = 5
at t = pi, x = 5 again, but the argument of the cos function is now 2pi, a movement of one cycle, or 4 times the amplitude
.: movement, 12 units, = 4 times B
B = 3
====

Using again the expression,

At t=0, x=5 = B + R/Q

5 = 3 + R/4
2 = R/4
R = 8
====
 
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Fermat:
|B| = 6 does not imply that B=6.
There is no "choice" of Q - read the situation.

While you're at it, read the sticky post:

please DO NOT do someone's homework for them or post complete solutions to problems. Please give all the help you can, but DO NOT simply do the problem yourself and post the solution (at least not until the original poster has tried his/her very best).

Members who:

(1) attempt to get others to do their work for them, or
(2) post the same question in multiple forums, or
(3) prematurely post complete solutions to problems,

are violating the spirit (and policy) of PF!
 
Hello lightrav,

Reading those stickies was among the first things I did when I came to this board.

If you read my earlier posts, you will see that I do try to adhere to the spirit (and policy) of this board.
 
OP has done so much work already, and shown it, so I don't see that Fermat has done too much for him.
 
Thanks a million folks :!) - I didn't ask for the entire solution, only pointers, but at least now I have a way of checking my work :cool: .
 
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