How Do Masses and Springs Affect Vibration Frequencies?

[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?

 

[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 $$l_{0}$$.

 

[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)