# Mechanical Vibrations

#### alane1994

##### Active member
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 $$u(0)=5$$?

Last edited:

#### MarkFL

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

$$\displaystyle u(0)=0$$

and take the positive direction to be up.

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

Gravity:

$$\displaystyle F_1=-mg$$

Restoring force (Hooke's Law):

$$\displaystyle F_2=-ku+mg$$

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:

$$\displaystyle \sum F=ma$$

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

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

Can you proceed?

#### alane1994

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

$$\displaystyle u(0)=0$$

and take the positive direction to be up.

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

Gravity:

$$\displaystyle F_1=-mg$$

Restoring force (Hooke's Law):

$$\displaystyle F_2=-ku+mg$$

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:

$$\displaystyle \sum F=ma$$

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

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

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 dunno why I am having such a hard time about grasping this.

#### alane1994

##### Active member
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

Staff member
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:

$$\displaystyle mg=kx$$

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

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

And on that note...I gotta run for a few hours.

#### alane1994

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

$$t=\dfrac{\pi}{14}s$$

#### MarkFL

Staff member
The general solution to the ODE is:

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

Hence:

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

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

$$\displaystyle u(0)=c_1=0$$

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

And so (in meters):

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

And thus (in centimeters):

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

Now, we see that:

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

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

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

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