Damped Oscillator Question?

[coffeem]

The equation for motion for a damped oscillator is:

x(double dot) + 2x(dot) + 2 = 0

a) Show that x(t)= (A + Bt)e^-t

Where A and B are constants, satisfies the equation for motion given above.

b) At time t = 0, the oscillator is released at distance Ao from equilibrium and with a speed Uo towards the equilibrium position. Find A and B for these initial conditions.

c) Sketch the t-dpendence of x for the case in which Ao = 20m and Uo =25m/s and the case in which Ao = 20m and Uo =10m/s.


MY ATTEMPT AT ANSWER

a) Can do fine. No probems with this.

b) Setting t = 0 gives x = A

So I am assuming as x = Ao then A - Ao.

However I do not know how to get further than this.

c) Dont know how to do this. Am assuming that once you have the relationships between Ao, Uo, A and B then you will be able to just plug the numbers in and graph the function.


Thanks for any help.

 

[alphysicist]

Hi cofeem,

You found that A=Ao by setting t=0 in the x(t) expression and knowing that it must equal Ao.

The other initial condition deals with the velocity. Since you know x(t), how do you find v(t)? What do you get? Then you can do the same thing with v(t) to find B that you did with x(t) to find A.

 

[Moetasim]

I can provide some guidance on how to approach this problem.

a) To show that x(t) = (A + Bt)e^-t satisfies the given equation for motion, we must substitute it into the equation and show that it equals 0. So, let's start with the left side of the equation:

x(double dot) + 2x(dot) + 2 = [(A + Bt)e^-t]'' + 2[(A + Bt)e^-t]' + 2

Using the product rule and chain rule, we can simplify this expression to:

[(A + Bt)'' - 2(A + Bt)' + 2(A + Bt)]e^-t

Now, we need to show that this equals 0. Recall that (A + Bt)'' = 0, since the second derivative of a linear function is 0. Also, (A + Bt)' = B, since the derivative of a constant is 0 and the derivative of Bt is B. So, our expression becomes:

[0 - 2B + 2(A + Bt)]e^-t = (2A - 2B + 2B)e^-t = 2Ae^-t

Since A is a constant, we can bring it outside of the parentheses:

2Ae^-t = 2Ae^-t

As you can see, the left side of the equation equals the right side, so x(t) = (A + Bt)e^-t does indeed satisfy the equation for motion.

b) To find A and B for the given initial conditions, we can use the fact that x(0) = A and x'(0) = B. So, for Ao = 20m and Uo = 25m/s, we have:

x(0) = A = 20m
x'(0) = B = 25m/s

And for Ao = 20m and Uo = 10m/s, we have:

x(0) = A = 20m
x'(0) = B = 10m/s

c) To sketch the t-dependence of x for the two given cases, we can use the equation x(t) = (A + Bt)e^-t and plug in the values of A and B that we found in