Finding The Position of A Mass As A Function of Time

[embphysics]

Homework Statement

Consider a mass m on the end of a spring of force constant k and constrained
to move along the horizontal x axis. If we place the origin at the spring’s equilibrium
position, the potential energy is U = 1/2kx^2 At time t = 0 the mass is sitting at the
origin and is given a sudden kick to the right so that it moves out to a maximum
displacement x_max = A and then continues to oscillate about the origin.

(a) Write down the equation for conservation of energy and solve it to give the
mass’s velocity \dot{x} in terms of the position x and the total energy E.

(b) Show that E = 1/2kA^2 and use this to eliminate E from your expression for \dot{x}. Find the time for the mass to move from the origin out to a position x.

(c) Solve the result of part (b) to give x as a function of t and show that the mass
executes simple harmonic motion with period 2 \pi \sqrt{m/k}

Homework Equations

The Attempt at a Solution

All right. I am currently trying to evaluate the integral, which was instructed of me to find in part (b).

Before I proceed any further, here is a little preliminary work:

\dot{x}(x) = kA^2 - \frac{k}{m} x^2

And I used this fact to find the integral: \dot{x} = \frac{dx}{dt} which becomes by separation of variables dt = \frac{dx}{\dot{x}}

Substituting in, and integrating both sides, this is the integral thus far:

\int_0^t dt' = \int_0 ^x \frac{dx}{\sqrt{kA^2 - \frac{k}{m}x^2 }}

kA^2 = (k^{1/2}A)^2 and \frac{k}{m}x^2= (\omega x)^2, where

\omega = \sqrt{\frac{k}{m}}. This leads to trigonometric substitution:

sin \theta = \frac{(k^{1/2}A)^2}{\sqrt{kA^2 -(\omega x)^2}}

The problem I have is finding how dx is equal to dθ. I tried using the fact that \cos \theta = \frac{( \omega x)^2}{(k^{1/k} A)^2}, then taking the arccos of this, and then taking the derivative of theta with respect to x. This didn't really seem helpful.

How can I get rid of dx, so that I might have an integral involving only theta?

 

[Simon Bridge]

All right. I am currently trying to evaluate the integral, which was instructed of me to find in part (b).

(b) Show that E = 1/2kA^2 and use this to eliminate E from your expression for ##\dot{x}. Find the time for the mass to move from the origin out to a position x.

Starting with:
$$\frac{dx}{dt}=kA^2+\omega^2 x^2 : \omega^2=\frac{k}{m}$$ I get $$dt = \frac{dx}{kA^2+\omega x^2}$$

 

[embphysics]

Are you certain that that is correct? Here is my work from the beginning:

E = T(\dot{x}) U(x) ~ \rightarrow E =\frac{1}{2}m \dot{x} + \frac{1}{2}kx^2

When x_max = A, then \dot{x} = 0; thus, E = 0 + \frac{1}{2}kA^2

Taking the equation and solving for \dot{x}:

\frac{1}{2}kA^2 =\frac{1}{2}m \dot{x} + \frac{1}{2}kx^2

\dot{x}^2 = kA^2 - \omega x^2

\dot{x} = \sqrt{kA^2 - \omega x^2}

And then I applied the fact that \dot{x} = \frac{dx}{dt}, which, by separation of variables, can be written as dt = \frac{dx}{\dot{x}}

I am not certain as to how you were able to do this without introducing a square root.

 

[Simon Bridge]

From post #1:

Before I proceed any further, here is a little preliminary work:
\dot{x}(x) = kA^2 - \frac{k}{m} x^2
And I used this fact to find the integral

... that's what I started from. I take it that's a typo?

\dot{x} = \sqrt{kA^2 - \omega x^2}

... makes more sense :) ... shouldn't that be $\omega^2$ though?

$$\frac{dx}{dt}= \sqrt{kA^2 - \omega^2 x^2}$$

Try substitution:
$$x=\sqrt{\frac{kA^2}{\omega^2}}\sin\theta\\

dx = \sqrt{\frac{kA^2}{\omega^2}}\cos\theta\; d\theta$$

 

[embphysics]

Yes, sorry for the typos. How where you able to get x = \sqrt{\frac{kA^2}{\omega^2}} \sin \theta?

 

[Simon Bridge]

Did you try the substitution?

To understand the choice of substitution, consider:
If the stuff under the surd was $1-x^2$ ... what would be the substitution?

 

[embphysics]

Oh, I just realized that I drew my triangle incorrectly. I wrote the square of the lengths of the hypotenuse and one of the legs; that is, for the hypotenuse, I wrote(k^{1/2} A)^2, when I should have written k^{1/2} A, and for the leg I wrote(\omega A)^2, rather than \omega A.

From the way I have my triangle drawn, \sec \theta = \frac{k^{1\2}A}{\sqrt{(k^{1/2}A)^2 - (\omega A)^2}} \implies \frac{\sec \theta}{k^{1/2} A} = \frac{1}{\sqrt{(k^{1/2}A)^2 - (\omega A)^2}}

For the differential of x, x = \frac{k^{1/2}A}{\omega} \sin \theta \implies dx = \frac{k^{1/2}A}{\omega} \cos \theta d \theta

Using these two expressions, I'll substitute them into the integral, for which I will get:

t = \int \frac{\sec \theta}{k^{1/2} A} \frac{k^{1/2}A}{\omega} \cos \theta d \theta

Simplifying,

t = \frac{1}{\omega} \int d \theta \implies t = \frac{\theta}{\omega}

Going back to the triangle, we find that theta is \theta = \sin^{-1} (\frac{\omega x}{k^{1/2}A})

Substituting this in, simplifying, and solving for x, I get

x = \sqrt{m}A \sin (\omega t).

How do I get rid of that square root of the mass?

 

[Simon Bridge]

Your dimensions don't stack up - check your algebra. (Dimensional analysis o the DE you constructed.)
Note: I can no longer see what you have done.