# Frequency of oscillation (spring)

1. May 18, 2011

### Gold3nlily

1. The problem statement, all variables and given/known data
Two identical springs of spring constant 240 N/m are attached to each side of a block of mass 21 kg. The block is set oscillating on the frictionless floor. What is the frequency (in Hz) of oscillation?

Figure:
http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c15/fig15_30.gif

2. Relevant equations
w = sq(k/m)
w = omega = angular frequency
k = spring constant
m = mass

w=2(pi)f
f = w/(2pi)

3. The attempt at a solution

So this problem seemed simple enough...
w = sq(k/m)
w = sq(240/21)
w = 3.38

f = w/(2pi)
f = 3.38/(2pi) = 0.538Hz

Why is this wrong? It must have something to do with there being two springs... but I don't know how that would change things.

2. May 18, 2011

### Staff: Mentor

Suppose that if the block were at rest at the equilibrium point you were to displace it, say, towards the left-hand spring by a small amount x. Each spring is going to react with some force. What would be the net force required to make the displacement? How does that compare to what a single spring would do?

3. May 18, 2011

### Gold3nlily

This may sound silly but wouldn't they cancel each other out becasue they are pulling from opposite directions?

hmm...

One would be pushing (KE?), and one would be pulling (PE?).
Maybe I am to use energy conservation?
There are no external forces, so then Emec is conserved.

KE = 1/2* K (Xm)2 sin2(wt + :theta:)
U(t)= 1/2* k (Xm)2cos2(wt + :theta:)

but it would be difficult to get "w" out of those functions. Is there a better way to do this?

4. May 18, 2011

### Staff: Mentor

One will be pulling and one will be pushing, yes. But pay attention to the directions of the forces that result! If the displacement x is to the left the spring on the left, being compressed, will push to the right. Meanwhile, the spring on the right, being stretched, will pull to the right. So both forces are to the right, opposite the direction of the displacement.

5. May 18, 2011

### Gold3nlily

Okay, so my guess is that I look at hook's law: F= -Kd
If the forces are moving together then -->
2F = -kd
so K will be half the size.

So then I apply that to this equation:
w = sq(.5k/21) = 2.39
f = w/(2pi) = (2.39/(2pi)) = 0.380 s-1??

6. May 18, 2011

### Staff: Mentor

No, the force is doubled. F = -2Kd.

What, then, is the effective spring constant?