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

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Homework Help Overview

The discussion revolves around solving a particle's differential equation of motion, specifically \(\frac {d^2x} {dt^2} = -Qx + R\). The original poster seeks to express the position \(x\) in terms of time \(t\), given specific initial conditions and constraints regarding the particle's motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve the differential equation and expresses \(x\) in terms of \(t\) using characteristic equations. Some participants question the assumptions regarding the dimensionality of the problem and the implications of the initial conditions on the constants \(Q\) and \(R\).

Discussion Status

Participants are actively engaging with the problem, exploring various interpretations of the equations and conditions presented. Some have provided insights into the relationships between the constants and the motion of the particle, while others emphasize the importance of adhering to the forum's guidelines regarding homework help.

Contextual Notes

There are constraints regarding the original poster's understanding and the need to avoid providing complete solutions. The discussion reflects a mix of attempts to clarify the problem and to ensure that the original poster is guided without being given direct answers.

SplinterIon
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Given a particle whose equation of motion is

[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.
 
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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=?
 
Last edited:
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
====
 
Last edited:
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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