Thought I'd nailed this What's gone wrong?

[TaliskerBA]

Homework Statement

The displacement x (in metres) of a damped pendulum from the vertical satisfies
x'' + x' + 10x = 0.

The pendulum is displaced 2cm from the vertical and released so that its initial
velocity is 0. Find the displacement of the pendulum to the other side at the end of
its first swing.

Homework Equations

x = exp(-0.5t)*(Pcos(t*sqrt(39)) + Qsin(t*sqrt(39)) (n.b. will now denote sqrt(39) as 'u')
x' = exp(-0.5t)*((uQ - 0.5P)cos(ut) - (uP + 0.5Q)sin(ut))

Initial conditions are x(0) = 0.02 and x'(0) = 0

The Attempt at a Solution

Using the initial conditions I get P=0.02 and Q = (100u)^(-1) which gives:

x = exp(-0.5t)*(0.02*cos(ut) + (100u)^(-1)*sin(ut))

Due to the exponential multiple part going to 0 as t goes to infinity the maximum/minimum oscillations get smaller over time. At t=0 the displacement is at its maximum (due to the damping effect) so my thinking is I want to find the next value of t at which x' = 0 to get the time at which the oscillation is at its minimum. First I differentiate x:

x' = - exp(-0.5t)*(156/200u)sin(ut)

Here I confirm that when t = 0, x'=0. Now I want to find the next point when x'=0, which is the next time sin(ut) = 0. I use sin(pi)=0:

pi = ut ---> t = pi/u = pi/sqrt(39)

When I feed this back into x I get x = -0.01555222255 which seemed to me to be a reasonable result and the displacement to the other side was approx 1.56cm after the first swing. The answer given in my tutorial notes is 2exp(-pi/sqrt(39)) = 1.209 which is positive and a lot larger than my answer.

This is one of the hard 'try it if you dare' kind of questions from my class so I might well have approached it completely wrong. Would appreciate some help and explanation as to what has happened here.

Many thanks.

 

[Dick]

I'm going to agree with the books answer if the units on it are centimeters. Which I think it is. The first place I start disagreeing with you is that I think the arguments in the sin and cos of x are sqrt(39)*t/2. Nothing wrong with your overall approach though.

 

[TaliskerBA]

Ah what a stupid mistake to make. I did everything right apart from that bit. It doesn't matter if I use initial conditions x(0) = 2 or x(0) = 0.02 as when I use the latter I now get 0.01209 which is the correct answer. Thanks for your help that would have bugged me.