MHB When does the mass first return to its equilibrium position?

[alane1994]

Here is my problem verbatim.

A mass weighing 100g stretches a spring 5cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10cm/s, and if there is no damping, determine the position \(u\) of the mass at any time \(t\). When does the mass first return to its equilibrium position?

For this, these are the things that I have been able to determine:

\(m=100~\text{grams}\)

\(\gamma=0\)

And I believe that we would use Newton's Law?

\(mu^{\prime\prime}(t)+\gamma u^{\prime}(t)+ku(t)=F(t)\)

And we would need initial conditions right?

\(u(0)=~?\\
u^{\prime}(0)=-10cm/s\)

I am rather stumped...EDIT:
Would \(u(0)=5\)?

 

[MarkFL]

I would orient my coordinate axis such that equilibrium is at:

$$u(0)=0$$

and take the positive direction to be up.

Now, you need to consider the forces acting on the mass:

Gravity:

$$F_1=-mg$$

Restoring force (Hooke's Law):

$$F_2=-ku+mg$$

Notice that when the mass is at equilibrium then the spring exerts a restoring force on the mass equal in magnitude but opposite in direction to the weight of the mass.

Now apply Newton's second law:

$$\sum F=ma$$

$$-ku=m\frac{d^2u}{dt^2}$$

$$\frac{d^2u}{dt^2}+\frac{k}{m}u=0$$

Can you proceed?

 

[alane1994]

I would orient my coordinate axis such that equilibrium is at:

$$u(0)=0$$

and take the positive direction to be up.

Now, you need to consider the forces acting on the mass:

Gravity:

$$F_1=-mg$$

Restoring force (Hooke's Law):

$$F_2=-ku+mg$$

Notice that when the mass is at equilibrium then the spring exerts a restoring force on the mass equal in magnitude but opposite in direction to the weight of the mass.

Now apply Newton's second law:

$$\sum F=ma$$

$$-ku=m\frac{d^2u}{dt^2}$$

$$\frac{d^2u}{dt^2}+\frac{k}{m}u=0$$

Can you proceed?

I am looking at this, and it is quite different than the examples in the book... and I would need to solve for \(k\) right? How would I do that? I would need the displacement \(L\) of the mass from equilibrium, right? I am sorry if I seem stupid right now. I don't know why I am having such a hard time about grasping this.:(

 

[alane1994]

Here is a picture from my teachers powerpoint slide.

And some text from the slide.
–Weight: w = mg (downward force)
–Spring force: Fs = - k(L+ u) (up or down force, see next slide)
–Damping force: Fd(t) = - g u'(t) (up or down, see following slide)
–External force: F (t) (up or down force, see text)

I guess where I am getting hung up, is that in all of the examples there is some sort of further displacement from its' equilibrium...

So for this problem I think \(L\) is 5 cm right?

 

[MarkFL]

To find $\dfrac{k}{m}$, we observe that the 100 gram (0.1 kg) mass stretches the spring 5 centimeters (0.05 m). Using Hooke's Law, we have:

$$mg=kx$$

$$\frac{k}{m}=\frac{g}{x}=\frac{9.8\frac{\text{m}}{ \text{s}^2}}{\frac{1}{20} \text{ m}}=196\frac{1}{\text{s}^2}$$

So, your IVP becomes:

$$\frac{d^2u}{dt^2}+196u=0$$ where $$u(0)=0\text{ m},\,u'(0)=\frac{1}{10}\frac{\text{m}}{\text{s}}$$

And on that note...I got to run for a few hours. :D

 

[alane1994]

\(u=\frac{5}{7}\sin(14t)~cm\)
\(\text{t is in seconds}\)

\(t=\dfrac{\pi}{14}s\)

 

[MarkFL]

The general solution to the ODE is:

$$u(t)=c_1\cos(14t)+c_2\sin(14t)$$

Hence:

$$u'(t)=14c_2\cos(14t)-14c_1\sin(14t)$$

Using the initial values (I should have stated earlier $$u'(0)=-\frac{1}{10}\frac{\text{m}}{\text{s}}$$ since the initial velocity is downward), we find:

$$u(0)=c_1=0$$

$$u'(0)=14c_2=-\frac{1}{10}\,\therefore\,c_2=-\frac{1}{140}$$

And so (in meters):

$$u(t)=-\frac{1}{140}\sin(14t)$$

And thus (in centimeters):

$$u(t)=-\frac{5}{7}\sin(14t)$$

Now, we see that:

$$u(t)=0\implies 14t=k\pi\implies t=\frac{k\pi}{14}$$

For $0<t$, the smallest value is then for $$k=1$$ or:

$$t=\frac{\pi}{14}\text{ s}$$

You did everything correctly, I threw you off by giving you the wrong direction for the initial velocity. (Doh)