1. ## Help! Mechanical vibrations.

I was wondering if anyone would be able to start me off on this question;

A Weight stretches a spring 3 inches. It is set in motion at a point 4 inches below it's equilibrium position with zero velocity.
-Find the max amplitude
-Find the max velocity
-When does it reach it's highest point

2. ## Simple Harmonic Motion

Hello s7b
Originally Posted by s7b
I was wondering if anyone would be able to start me off on this question;

A Weight stretches a spring 3 inches. It is set in motion at a point 4 inches below it's equilibrium position with zero velocity.
-Find the max amplitude
-Find the max velocity
-When does it reach it's highest point
Assuming that Hooke's Law is still obeyed when the spring becomes compressed, the motion is Simple Harmonic, with centre the original equilibrium position, and amplitude 4 inches. That's the answer to the first question.

The tension in the spring is proportional to the extension. When the extension is 3 inches, tension = weight of body = mg. So when extension = 7 inches, tension = $\displaystyle \frac{7mg}{3}$. So, if we resolve vertically at the point of maximum extension:

$\displaystyle mg - \frac{7mg}{3} = m \times acceleration$

$\displaystyle \Rightarrow acceleration = -\frac{4g}{3}= -\frac{128}{3}$, taking g = 32

With the usual notation, the equation of motion is

$\displaystyle \ddot{x} = -\omega^2 x$

And we know that when $\displaystyle x = \frac{1}{3}$ feet, $\displaystyle \ddot{x} = -\frac{128}{3}$. So we can now work out the value of $\displaystyle \omega$

Now the velocity, displacement and amplitude are related by

$\displaystyle v^2 = \omega^2(a^2 - x^2)$

The maximum velocity is when $\displaystyle x=0$. So max velocity = $\displaystyle a\omega$. And the time it takes to reach its highest point is half the period of a complete oscillation. And that period is

$\displaystyle \frac{2\pi}{\omega}$

Can you take it from here?

3. where did you get the 7??

4. Also, is there a way of setting up a differential equation anf solving it that way??

5. ## Simple Harmonic Motion

Hello s7b
Originally Posted by s7b
where did you get the 7??
In the equilibrium position, the extension of the spring is 3 inches. It's then extended a further 4 inches. 3 + 4 = 7.
Originally Posted by s7b
Also, is there a way of setting up a differential equation anf solving it that way??
Yes, although I would expect that you would have covered all this if you have studied SHM.

But, from first principles, it goes something like this (and I'm not using particular measurements or particular units now):

Suppose that the original extension of the spring when the particle is in equilibrium is $\displaystyle b$. Then in this position, tension in spring = $\displaystyle mg$. So when the spring has a further extension $\displaystyle x$ below the equilibrium position, the tension, using Hooke's Law, is

$\displaystyle T = \frac{(b + x)mg}{b}$

So, in this position, use force = mass x acceleration:

$\displaystyle mg - T = m\ddot{x}$

$\displaystyle \Rightarrow mg - \frac{(b + x)mg}{b} = m\ddot{x}$

$\displaystyle \Rightarrow \frac{mgb - mgb - mgx}{b} = m\ddot{x}$

$\displaystyle \Rightarrow \ddot{x} = -\frac{g}{b}x$

This is the standard SHM equation, where $\displaystyle \omega^2 = \frac{g}{b}$. The solutions are well known - see for instance Simple Harmonic Motion -- from Wolfram MathWorld

But if you want to solve it for yourself, you could say that this is a second order differential equation, whose Auxiliary Equation is:

$\displaystyle m^2 + \frac{g}{b} = 0$

and solve it in the usual way to get $\displaystyle x$ in terms of $\displaystyle t$.

The solution takes the form $\displaystyle x = a\cos(\omega t + \alpha)$, where $\displaystyle a$ is the amplitude and $\displaystyle \alpha$, the phase angle, is determined by the initial conditions. If $\displaystyle x = a$ when $\displaystyle t = 0$ (as in our case), $\displaystyle \alpha = 0$, and we get

$\displaystyle x = a\cos(\omega t)$, where $\displaystyle \omega^2 = \frac{g}{b}$

The period of the oscillation is then $\displaystyle \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{b}{g}}$

Or you could write the acceleration as $\displaystyle v\frac{dv}{dx}$ and then solve

$\displaystyle v\frac{dv}{dx} = -\frac{g}{b}x$

to get

$\displaystyle \frac{1}{2}v^2 = -\frac{g}{2b}x^2 + c$

Then use the fact that, at the maximum displacement, $\displaystyle x = a$ (the amplitude) and $\displaystyle v = 0$. This gives

$\displaystyle v^2 = \frac{g}{b}(a^2 - x^2)$

... and so on.