Mass connected to two springs - simple harmonic motion

[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.

 

[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.
To help you a bit on your way:
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})
5. Use Newton's 3.law on point 3 to find the force acting on spring 2 from the mass.
Use the fact that spring 2 is massless to derive a relationship between L_{1}(t),x(t)
(That is, set up Newton's second law for spring 2.)
6. Set up Newton's 2.law of motion for the mass, and identify the frequency.

 

[wais]

To show that the period for this configuration is given by T = 2 \pi \sqrt{m ( k^{-1}_1 + k^{-1}_2)}, we can use the equation for the period of simple harmonic motion: T = 2 \pi \sqrt{\frac{m}{k}}.

In this case, the effective spring constant for the system can be found by combining the two spring constants in parallel, using the equation k_{eff} = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}}.

Substituting this into the equation for period, we get:

T = 2 \pi \sqrt{\frac{m}{k_{eff}}} = 2 \pi \sqrt{\frac{m}{\frac{1}{\frac{1}{k_1} + \frac{1}{k_2}}}}

Simplifying this, we get:

T = 2 \pi \sqrt{m (k_1 + k_2)}

Which is the same as:

T = 2 \pi \sqrt{m ( k^{-1}_1 + k^{-1}_2)}

Therefore, we have shown that the period for this configuration is indeed given by T = 2 \pi \sqrt{m ( k^{-1}_1 + k^{-1}_2)}. This means that the period of the system is dependent on the mass of the object and the combined spring constant of the two springs, which makes intuitive sense as both of these factors affect the force and thus the motion of the mass.