Find a function of velocity relative to x

[etchzel]

Homework Statement

An object is moving in x-axis with given force $Fx = -mω^{2}x$, where ω is a positive constant. when t = 0 the object starts moving from point $x_{0} > 0$.
a. Find the object's velocity function relative to x ($v(x)$)
b. Find the object's position function relative to t ($x(t)$)

Help: $\int\frac{dz}{\sqrt{A^{2} - z^{2}}}$ perform the integration by changing z = A cos u
(I don't know what this help thing for though, I find it irrelevant to the question )

Homework Equations

The Attempt at a Solution

a. First, using the above-mentioned force equation, I found the acceleration function relative to x, it goes like:

$Fx = -mω^{2}x$
$ma(x) = -mω^{2}x$
$a(x) = -ω^{2}x$​

then, since $a(x) = \frac{v(x)}{dt}$, by using chain rule, it goes:

$a(x) = \frac{v(x)}{dx}\frac{dx}{dt}$
$a(x) dx = v(t) dv(x)$​

assuming that v(t) = v(x) (even though the function might be different, the result will be the same either way) then:

$a(x) dx = v(x) dv(x)$
$\int a(x) dx = \int v(x) dv(x)$
$-\frac{1}{2}ω^{2}x^{2} + C = \frac{v(x)^{2}}{2} + C$
$v(x) = \sqrt{-1}ωx$​

there goes my attempt for part a, and I'm not even sure if it was correct, that $\sqrt{-1}$ part is throwing me off

for part b, my last-moment attempt goes like this:

$v(t) = \int a(x) dt$
$v(t) = -ω^{2}xt + C$​

when t = 0, the object is at rest, then v(0) = 0, and when t = 0, $x = x_{0}$

$0 = -ω^{2}x_{0}(0) + C ==> C = 0$
$v(t) = -ω^{2}xt$​

then using integral to find x(t):

$x(t) = \int v(t) dt$
$x(t) = \frac{-ω^{2}x_{0}t^{2}}{2} + C$​

when t = 0,

$x_{0} = \frac{-ω^{2}x_{0}(0)^{2}}{2} + C ==> C = 0$
$x(t) = \frac{-ω^{2}xt^{2}}{2} + x_{0}$​

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pretty sure I was wrong for part b :s
can someone give me a hint on how to solve this problem?
any help is much appreciated, thanks!

 

[cepheid]

Welcome to PF!

then, since $a(x) = \frac{v(x)}{dt}$, by using chain rule, it goes:

$a(x) = \frac{v(x)}{dx}\frac{dx}{dt}$
$a(x) dx = v(t) dv(x)$​

What exactly are you doing? I'm afraid this is nonsense. The acceleration is the time derivative of the velocity. So, it is:$$a(x) = \frac{dv(x)}{dt}$$and so by the chain rule:$$a(x) = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$$

 

[cepheid]

Now thing to do is to look at the right hand side of that last equation, and notice that it looks like something that *has already been* differentiated with respect to x using the chain rule. So, can you reverse that step, thus turn the right hand side into an exact derivative of something? I.e., right hand side can be expressed as $\frac{d}{dx}[\textrm{blah}]$, and you have to figure out what "blah" is.