## Oscillation with Two Springs (SMH)

In the figure below, two springs are attached to a block that can oscillate over a floor. If the left spring is removed, the block oscillates at a frequency of 27 Hz. If, instead, the spring on the right is removed, the block oscillates at a frequency of 43 Hz. At what frequency does the block oscillate with both springs attached?

So I know that f=(1/2pi)*sqrt(k/m)
k=((2*pi*f)^2)

Using these formulas, I solved for each coefficient of friction. Then, I used 1/k= 1/k1 +1/k2 to find the total k value. I plugged that into the frequency equation and got 22.866 Hz. However, webassign tells me that's wrong. Any ideas on where I screwed up? I only have one try left.
 Anybody? I'm completely stuck here..
 The way you combine the spring constants is not right for this case. It works if the two springs are connected to each other and then the mass is at the end of one of the springs. Here is a different situation. Think about what happens when the object is displaced by $$\Delta x$$ (either to the left or to the right). What will be the total elastic force on the object? If you answer this question, you will know the "equivalent" elastic constant.

## Oscillation with Two Springs (SMH)

So.. would the force be F= -k1(x)-(-k2)x?

I'm not really sure how to approach the problem this way..
 Nevermind, I figured it out! Thanks a lot for the help!

 Quote by Dante Tufano So.. would the force be F= -k1(x)-(-k2)x? I'm not really sure how to approach the problem this way..
The elastic forces are in the same direction. The force is (k1+k2)x (or minus this if you consider the sign.

 Tags harmonic, motion, oscillation, simple, springs