block 'n springs

**billym** · April 1st 2008, 04:01 PM

A block of mass $m$ lies on a smooth plane $AB$ inclined at an angle $\alpha$ to the horizontal. The block is attached to $A$ and $B$ by a spring on either side (up the slope is positive).

The general solution for the equation of motion is:

$ x\left ( t \right ) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right ) $

Given:

$ x\left ( 0 \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) $

I have the particular solution as:

$ x\left ( t \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) cos \left ( t \sqrt{\frac{5k}{m}} \right ) $

The period as:

$ \frac{2\pi}{\sqrt{\frac{5k}{m}}} $

and the amplitude as:

$ l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) $

seems weird... is all this correct?

Is $\phi = \frac{dx}{dt} = 0$ ?

If so, would the average position of the block be the equilibrium position?

**topsquark** · April 2nd 2008, 10:21 AM

Originally Posted by billym

A block of mass $m$ lies on a smooth plane $AB$ inclined at an angle $\alpha$ to the horizontal. The block is attached to $A$ and $B$ by a spring on either side (up the slope is positive).

The general solution for the equation of motion is:

$ x\left ( t \right ) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right ) $

Given:

$ x\left ( 0 \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) $

I have the particular solution as:

$ x\left ( t \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) cos \left ( t \sqrt{\frac{5k}{m}} \right ) $

These last two lines are confusing me.

If
$x(t) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right )$

the only way that you can possibly satisfy your initial condition with the solution
$ x\left ( t \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) cos \left ( t \sqrt{\frac{5k}{m}} \right ) $
is if you aren't using parenthesis.

Please verify these two lines.

-Dan

**topsquark** · April 2nd 2008, 10:55 AM

Ahhhh! Success!

I want to state first that I do not understand the Physics of this situation. I have tried to "reverse engineer" the problem and determine spring constants, initial positions etc, from the x(t) equation and the initial conditions and have come up with conflicting facts. Perhaps there is just something here I'm not seeing, but I would double check to make sure your general solution here is valid.

I see what has happened here. You didn't list all of your iniitial conditions. You needed to state the condition:
$v(0) = 0$

So what we have here is a general solution
$x(t) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right )$

with
$x(0) = l_0 - \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha )$
and
$v(0) = 0$

For convenience, I'm going to define
$\omega = \sqrt{\frac{5k}{m}}$
and
$L = \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha )$

So your problem becomes:
$x(t) = A~cos( \omega t + \phi )$
with
$x(0) = l_0 - L$
and
$v(0) = 0$

Let's work with the velocity condition first:
$v(t) = \frac{dx}{dt} = -A \omega ~sin( \omega t + \phi ) = 0$

For t = 0:
$-A \omega ~sin(\phi) = 0$

Thus $\phi = 0, \pi$ . Which one we choose is going to depend on which spring is exerting the greater force. You have not given us that information. I will show you how the equation shows this factor.

So the solution will take the form of:
$x(t) = A~cos( \omega t )$ or $x(t) = A~cos( \omega t + \pi )$
with
$x(0) = l_0 - L$

Now, $cos( \omega t + \pi ) = cos( \omega t )~cos( \pi ) - sin( \omega t)~sin(\pi) = -cos(\omega t)$

Thus the solution is
$x(t) = \pm A~cos( \omega t )$
with
$x(0) = l_0 - L$
Which one of the + or - depends on the phase angle.

So to continue:
$x(0) = \pm A~cos(0) = \pm A = l_0 - L$

The size of L is going to determine if A is positive or negative. I leave it to you to decide if $l_0 - L$ can be negative or not. (The general solution for x(t) seems to imply one thing and the initial condition seems to imply another so I cannot reliably predict the size of L.)

I will simply leave this as
$A = \pm (l_0 - L)$

As a curious consequence of the initial condition notice that
$x(t) = \pm ( \pm (l_0 - L)~cos( \omega t)$
The $\pm$ signs are taken to be in the same sense, so they cancel out, leaving
$x(t) = (l_0 - L)~cos( \omega t)$
which, when put into the form of the original variables leads to

$x(t) = \left [ l_0 - \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha ) \right ] ~ cos \left ( t \sqrt{ \frac{5k}{m}} \right )$

(We also see that I was right: You didn't use parenthesis in your original post, as this is obviously what you had meant to write.)

As to the rest, yes, the amplitude is indeed
$l_0 - \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha )$

But $\phi$ has nothing at all to do with dx/dt = v. I don't know why you would say that. This is not true, if for no other line of reasoning, because the units are wrong. And besides
$v(t) = \frac{dx}{dt} \neq 0$
in general as you can see by taking the derivative of x(t).

-Dan

**billym** · April 2nd 2008, 11:22 AM

Whoops, sorry...

Background:

The first spring attached from $A$ to the block has stiffness $2k$ , and natural length $l_0$ .

At time $t$ , the displacement of the block from $A$ , measured along the slope is $x$ .

The second spring attached from $B$ to the block has stiffness $3k$ and natural length $2l_0$ .

The distance between $A$ and $B$ is $5l_0$ .

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