# Homework Help: Spring-mass system in an excited box

1. Jun 1, 2012

### PhMichael

1. The problem statement, all variables and given/known data

Well, I have this spring (stiffness $k$ and free length $l_{0}$) mass ($m$ ) system in a box which has a width $w$ such that $l_{0}>w$ (i.e. the spring is compressed). The box is excited (given a prescribed position) by: $u(t)=b\cdot sin({\omega}\cdot t)$. Determine the range of possible frequencies $\omega$ for which the mass does NOT lose contact with the right wall of the box.

Answer: $${\omega} < \sqrt{\frac{l_{0}-w}{b}\cdot{\frac{k}{m}}}$$

My solution:

The acceleration of the mass is:
$$a=-b\cdot \omega^{2} \cdot sin(\omega \cdot t)$$

Therefore,

$$-N -k \cdot (b\cdot sin(\omega \cdot t) + l_{0} - w) = -m b \cdot \omega^{2} \cdot sin(\omega \cdot t)$$

Now, if I isolate $N$ and require that $N>0$, I don't get to that answer. In fact, my answer will obviously depend on this sine function too. What am I doing wrong? what is the correct approach for solving this question?

Thanks!

2. Jun 2, 2012

### rude man

I would:

assume the box has an open right-hand-side. So now you just have a spring with a mass at one end and an excitation function at the other. Write down the equation and solve for x(t) (let x = 0 at left-hand side of box when bsin(ωt) = 0). x(t) is the position of m. So w is not included in this equation.

So now you just compare x(t) with bsin(ωt) + w which is the position of the right-hand side of the box. ω then has to be such that
x(t) >= bsin(ωt) + w for all t.

3. Jun 2, 2012

### Dickfore

If the mass does not lose contact with the wall, then it moves the same as the wall, which means its acceleration is also the same. The forces acting on the wall are the elastic force from the compressed string acting to the right, and the force of normal reaction acting to the left. The condition that the mass stays in contact with the wall is that the normal reaction force remains non-negative.

4. Jun 2, 2012

### PhMichael

$$-k\cdot{(x-b\cdot{sin(\omega \cdot t)}-l_{0})}=-m\cdot b \cdot \omega^{2} sin(\omega \cdot t) \Rightarrow \boxed{x = \frac {m}{k} \cdot b \cdot \omega^{2} \cdot sin(\omega \cdot t) + b\cdot sin(\omega \cdot t)+l_{0}}$$
Now,

$$\frac {m}{k} \cdot b \cdot \omega^{2} \cdot sin(\omega \cdot t) + b\cdot sin(\omega \cdot t)+l_{0} \geq b \cdot sin(\omega \cdot t) + w \Rightarrow \boxed{\omega^{2} \cdot sin(\omega \cdot t) \geq \frac{k}{m} \cdot \frac {w-l_{0}}{b}}$$

I don't get the answer (I posted the correct answer in my first post)... and it still depends on the sine function. What did I do wrong?

Well, this is exactly what I did in the first post and still got it wrong =/

5. Jun 2, 2012

### Dickfore

What's the red term supposed to be and is the direction of the force from the spring correct?

6. Jun 2, 2012

### PhMichael

The direction of the force from the spring should be "+" (to the right). However, I'm not sure what the answer of your first question.

The spring is compressed by an amount $l_{0} - w$ with respect to the box, however, I need to express it with respect to the fixed left end and thus, this red term appears. No?