Finding the Effective Spring Constant for Joined Springs

[apchemstudent]

Two springs with a spring constant of k = 6430 N/m are joined and connected to a block of mass 0.245 kg. The system is then set oscillating over a frictionless surface. What is the frequency of the oscillations?

This is what I think is the correct approach to this question:

since the springs are joined, the new spring now has a spring constant of 6430/2 = 3215 N/m.

So f = sqrt(k/m)/2*pi

= 18.23 Hz.

The springs are in series.
Is this correct? Thanks.
As well, how will I be able to solve this question if the 2 springs were not identical?

 

[robphy]

Start by drawing free-body diagrams. Then, consider properties to determine the "equivalent spring"... analogous to the problem of determining the "equivalent capacitance" in a capacitive circuit.

 

[apchemstudent]

Start by drawing free-body diagrams. Then, consider properties to determine the "equivalent spring"... analogous to the problem of determining the "equivalent capacitance" in a capacitive circuit.

So Equivalent spring is = (1/k1 + 1/k2 + 1/k3... 1/kn)^-1
where kn is the spring constant for each component ? This is based on what you said.

This is how I would approach it:

Let's say there are only 2 springs connected in series. We can calculate the ratio between the distances stretched by each component. Since the Tension force is constant, we can use those ratios to calculate the equivalent spring constant.

Ex. Ratio 4:1 stretched between springs A and B.

so to calculate the equiv spring constant, 4/5*Ka + 1/5*Kb = Equivalent spring constant.

This is what i think, but I'm not sure if it is correct.

 

[Tide]

Are the springs both on the same side of the mass or on opposite sides?

 

[Jelfish]

Remember that in the electro-mechanical analogy, the capacitance is analogous to the compliance, which is the reciprocal of the spring constant.

 

[apchemstudent]

Are the springs both on the same side of the mass or on opposite sides?

they are on the same side

 

[lightgrav]

The Tension is the same, and you add the stretch distances (s=T/k).
k_effective = T/s_total , so you end up with the first line in post #3.