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

In summary, 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 =
  • #1
SplinterIon
17
0
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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  • #2
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:
  • #3
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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  • #4
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!
 
  • #5
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.
 
  • #6
OP has done so much work already, and shown it, so I don't see that Fermat has done too much for him.
 
  • #7
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: .
 

1. What is the purpose of solving a particle's differential equation?

The purpose of solving a particle's differential equation is to determine the position and charge of the particle at a specific time. This can help us understand the behavior and movement of particles in various systems and environments.

2. How is the position (x) of a particle related to its charge (Q) and resistance (R)?

The position of a particle is related to its charge and resistance through the differential equation: d^2x/dt^2 = (Q/m)x + (Q/R)dx/dt. This equation shows the relationship between the acceleration of the particle, its charge, and the resistive forces acting upon it.

3. What is the process for solving a particle's differential equation?

The process for solving a particle's differential equation involves first identifying the initial conditions of the particle, such as its initial position and velocity. Then, using the differential equation and these initial conditions, we can solve for the position and charge of the particle at any given time using mathematical techniques such as integration.

4. Can the solution to a particle's differential equation be affected by external forces?

Yes, the solution to a particle's differential equation can be affected by external forces such as electric or magnetic fields. These forces can be incorporated into the differential equation and can alter the movement and behavior of the particle.

5. How can solving a particle's differential equation be applied in real-world situations?

Solving a particle's differential equation can be applied in various real-world situations, such as predicting the trajectory of a charged particle in an electric field, understanding the movement of particles in a fluid, or analyzing the behavior of particles in a circuit. It is a fundamental tool in many areas of science and engineering.

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