# Finding the velocity of a spring

1. Jun 30, 2015

### NooDota

1. The problem statement, all variables and given/known data

If I have a spring that repeats its motion in T0 =2s, and its x coordinate is given by x=00.4*Cos(ω0t + φ).

If I wanted the velocity at, let's say t=1/2s, I would just take the derivative of x and plug in t.

My question is, what other ways can I use to find V at t = 1/2s? The spring's length and mass are both known.

If I graph the position function on x, t axis, and then take the point which has coordinates (position at the wanted time, 1/2), and then take its y/x, would I get the velocity at that time? Since velocity is the slope, and slope is y/x?

2. Relevant equations

3. The attempt at a solution

2. Jun 30, 2015

### RUber

Slope is $\frac{dy}{dx}.$ If you graph the position vs. time, and draw the tangent line to the plot at the time you want, then you would be approximating the derivative. However, in this example, you seem to know everything you need in order to just evaluate the derivative.

If you took its y/x, by which I assume you mean t/x you might get something like .5/0 which is definitely not the velocity.

Another way to approximate the velocity would be to see that every 2 seconds, the cycle repeats, meaning that the velocity is zero at both ends of the cycle, or every 1 second. Assuming that your $\varphi=0$, t=1/2 sec will be halfway between these stops, so the spring should be at its max velocity for the cycle. Do you know what the maximum velocity of your spring is?

3. Jun 30, 2015

### NooDota

φ is π

ω0 is also π

Xmax is 0.04

(And at the required time, the velocity is at its max value)

And from the law Vmax= ω0 * Xmax I'll get Vmax to be +0.04π m.s^-1, which is also the same answer I get from using the typical derivative way. So that's a second way.

Is there a third way? I really don't care if it's complicated.

Last edited: Jun 30, 2015
4. Jun 30, 2015

### RUber

Starting from Hooke's Law that says F=-kX where F is force, k is a positive constant, and X is position (displacement from rest).
Noting that F = mA and A = $\frac{d}{dt^2}X$
This gives you mX'' = -kX, or X'' = -cX with c>0.
You are looking for X', which can be found with a little magic.
Multiplying both sides by X' and integrating:
$\int_0^T X'' X' dt = -c \int_0^T X X' dt \\ X'X'|_0^T - \int_0^T X' X'' dt =-c XX|_0^T + c\int_0^T X' X dt$
$X'X'|_0^T=-c XX|_0^T$
Assuming that $X'(0) = 0$, you have $X'(T) = \pm \sqrt{-c(X(T)^2 - X(0)^2 ) }$
Checking this against your problem, with $c = \omega^2$
This can be written as $X'(T) = \pm \omega \sqrt{ X(0)^2 - X(T)^2 }$
$X(0) = -.4, X(1/2) = 0$
so $X'(1/2) = \pm \sqrt{\pi^2(.4^2)} = \pm .4\pi$
where the + or - can be determined by which part of the cycle you are in. In the case of your problem, you are on the forward velocity leg during T=1/2.

Is this what you were looking for in terms of a third way?

5. Jun 30, 2015

### NooDota

Okay, lol, that is a little bit complicated for me, especially since they haven't taught us integration in school yet. I still understand like 80% of what you did. Thanks.

Except, why did you plug in 0.4 instead of 0.04?

6. Jun 30, 2015

### RUber

I must have misread your original post that said 00.4. I guess you meant 0.04. No problem.