Question

consider the unforced, undamped spring system, modelled by m(d^2x/dt^2)+kx=0, and take w=sqt(k/m). Given initial conditions x(0)=xo and x'(0)=vo not both zero, show that x(t)+Acos(wt-a) for A= sqrt(xo^2+(vo/w)^2) and a defined by cosa=xo/A, sina=vo/wA

Answer 1

I'm so confused, how am I supposed to show this?

Answer 2

\[x(t) = A\cos(\omega t - \alpha)\]

Answer 3

find the second derivative and plug it in the differential equation

Answer 4

Ohhhh... I was finding the derivative to the general solution lol I'll try that, thanks

Answer 5

I get (-mw^2+k)(Acos(wt-alpha) :S

Answer 6

right, which is same as :

\[(-m\omega^2 + k )x = 0\]

yes ?

Answer 7

solve \(\omega \)

Answer 8

Ohhh, same as what's given. What does the A and the alpha part have to do with it?

Answer 9

Okay, so do we agree x(t) = Acos(wt-alpha) satisfies the given differential equation if we assume w = sqrt(k/m) ?

Answer 10

Yes

Answer 11

plugin the initial conditions and find out A and alpha

Answer 12

\(x(0) = x_0 \)
\[x_0 = A\cos(0-\alpha)\]
\[x_0^2 = A^2\cos^2\alpha\tag {1}\]

Answer 13

get another equation using the other initial condition and solve A and alpha

Answer 14

why did you square it?

Answer 15

you will see it after getting the second equation

Answer 16

I got the second equation, I know that w is squared but I don't see how that helps.
I got \[x'(0)=-Aw^2\cos(-\alpha)=v _{o}\]

Answer 17

i think you're substituting in the wrong equation

Answer 18

and then \[A=\frac{ \cos^{-1}(-\alpha)v_o }{ w^2 }\]

Answer 19

looks you're using x''(t) instead of x'(t)

Answer 20

Oh yeah haha

Answer 21

So when you square it alpha doesn't square?

Answer 22

wait dumb question

Answer 23

Okay I still don't see it, \[x'(0)=-Awsin(\alpha)=v_o\]
are you squaring both sides then adding them together?

Answer 24

square it and eliminate alpha
because that what the problem wants

Answer 25

\[\sin^2\alpha = \left(\frac{v_0}{A\omega}\right)^2\]

\[\cos^2\alpha = 1-\left(\frac{v_0}{A\omega}\right)^2\tag{2}\]

Answer 26

solve (1) and (2) for A