Homework Help: Three Springs and Two Masses

1. Oct 5, 2004

e(ho0n3

Here is the situation: There is a spring connected to a wall at one end and a mass m1 at the other, which in turn is connected to another spring, which is connected to mass m2, which is connected to another spring which is connected to another wall. In other words:

Wall ----- m1 ----- m2 ----- Wall

where ----- represents a spring (all with spring constant k). Here are the questions:

(a) Apply Newton's 2. law to each mass and obtain two differential equations for the displacements x1 and x2.
(b) Determine the possible frequencies of vibration by assuming a solution of the form x1 = A1 cos ωt, x2 = A2 cos ωt.

(a) There are only two forces acting on the masses in the horizontal direction, namely the spring force, so I figured the equations are:
m1a1 = -2kx1
m2a2 = -2kx2

(b) The frequencies are:
$$f_1 = \frac{1}{2\pi}\sqrt{\frac{2k}{m_1}}$$

$$f_2 = \frac{1}{2\pi}\sqrt{\frac{2k}{m_2}}$$

Somehow I don't think it would be this easy. Maybe I'm missing something? What do you think?

2. Oct 6, 2004

BLaH!

Your equations of motion aren't quite correct. Remember, spring forces are given by

$$F_{spring} = -k(l_{0} - \Delta x$$)

where $$\Delta x$$ is the compression of the spring from its natural length $$l_{0}$$.

3. Oct 6, 2004

arildno

ehoon:
1.Let the positions of the masses (measured from the left-hand wall) be:
$$x_{1}(t), x_{2}(t) (x_{2}>x_{1})$$
2. Spring forces on mass 1.
a) Spring 1 attached to wall:
Clearly, a stretch of this spring will give a force in the negative direction, so:
$$F_{11}=-k(x_{1}-l_{0})$$, where $$l_{0}$$ is the rest length of the spring.
b) Spring 2 attached between the masses:
Clearly a stretch of this spring will impart a force in the positive direction on mass 1, so:
$$F_{12}=k(x_{2}-x_{1}-l_{0})$$
(I've assumed that the rest lengths are equal)
3. Spring forces on mass 2:
a) Spring 2 attached between the masses:
Clearly a stretch of this spring will impart a force in the negative direction on mass 2, so:
$$F_{22}=-k(x_{2}-x_{1}-l_{0})$$
b) Spring 3 attached to the right-hand wall:
Clearly a stretch of this spring will impart a force in the positive direction on mass 2, so:
$$F_{23}=k(L-x_{2}-l_{0})$$
where $$L=3l_{0}$$ is the distance between the walls.
Hence, you get the system of equations:
$$m_{1}\ddot{x}_{1}+2kx_{1}-kx_{2}=0$$
$$m_{2}\ddot{x}_{2}+2kx_{2}-kx_{1}=3kl_{0}$$

You might, of course, rewrite this system of equations in terms of displacements, rather than positions (not that I see much point in that)