00:01
So in this question, we're told that when a mass of 2 kilograms is attached to a spring whose constant is 32 newtons per meter, it comes to rest of the equilibrium position.
00:12
So x of 0 equals v of 0, and then starting at t equals 0, f of t is 102.
00:21
E to the minus 2t, cos 4t.
00:26
So let's think about the equation of motion.
00:29
Well, f equals ma.
00:33
On the left -hand side, we have the force, which is minus kx plus f of t.
00:40
On the right -hand side, we have m times the acceleration, so m x -double dot.
00:50
So we have mx double dot plus kx is equal to 1 .02 e to minus 2t, cos 4t.
01:04
So we need to look for solutions to this equation.
01:10
So let's look for, first of all, let's look for the homogeneous solutions.
01:16
So mxh double dot plus k xh is equal to zero.
01:23
This is going to be solved by xh is equal to, is going to be equal to a cos root k over m t plus delta, which is just standard simple harmonic oscillator.
01:46
And now we need a particular integral.
01:48
And a particular integral will satisfy the whole equation.
01:52
So what we can do is we can look for something of the form.
01:58
Try xp is something like m.
02:06
E to the minus 2t times cos4.
02:11
Plus n e to minus 2 t sign 40 then xp dot is going to be minus 2m a to minus 2 t cos 40 minus 4m e to minus 2 t sine 40 minus 2n e to the minus 2 t sine 40 minus 2 n e to minus 2t cos 40.
02:48
So that's collect up the causes and signs.
02:52
We've got 4n minus 2m, e to the minus 2t, cos 4t, plus minus 2n minus 4m, ether minus 2t, sine 4t.
03:10
So now let's take another derivative, xp double dot, is going to be, minus 2, 4n minus 2m, ether minus 2t, cos 4t, minus 4, 4n, minus 2m, ether minus 2 t, sine 40, and then minus 2, minus 2n, minus 4m, a to minus 2 t, sine 4 t, and then minus plus 4m, plus 4 minus 2n minus 4m, ether minus 2t, cos 40.
03:57
So let's collect this up.
03:59
So the cos is we have 8n minus 8n.
04:06
So there's no ends.
04:08
And then we have, sorry, this is minus 8n, minus 8n is minus 16n...