# Mass connected to two springs - simple harmonic motion

1. Nov 17, 2004

### Skomatth

A mass m is connected to two springs, with spring constants k1 and k2. (m connected to k2, which is connected to k1 which is connected to a wall horizontally).

Show that the period for the configuration is given by
$$T = 2 \pi \sqrt{m ( k^{-1}_1 + k^{-1}_2)}$$

Well I know .5kA^2 = .5mv^2
If I could figure out how the spring compresses when a certain amount of work is done on it I think I could figure out how to show what the period is, but I'm not sure how it works with 2 springs. If someone could tell me how this works or show me a different direction to go it would help.

2. Nov 18, 2004

### arildno

What you need to remember, is that we approximate both springs as MASSLESS.
An important consequence of this, is that there can be no net force acting on each spring, if all accelerations are to be finite.
1. Let $$L_{1},L_{2}$$ be the rest lengths of each spring.
2. Let $$L_{1}(t)$$ be the time dependent length of spring 1; x(t) the position of the mass; hence, the time-dependent length of spring 2 fulfills $$L_{2}(t)=x(t)-L_{1}(t)$$
3. The force acting on the mass from spring 2 is $$-k_{2}(L_{2}(t)-L_{2})$$
4. The force acting on spring 1 on spring 2 fulfills: $$-k_{1}(L_{1}(t)-L_{1})$$
Use the fact that spring 2 is massless to derive a relationship between $$L_{1}(t),x(t)$$