Frequency of oscillation (spring)

In summary, the problem involves two identical springs with a spring constant of 240 N/m attached to a block of mass 21 kg, set to oscillate on a frictionless floor. To find the frequency of oscillation, the equations w=sqrt(k/m) and f=w/(2pi) are used. However, due to the displacement caused by the two springs pulling and pushing in opposite directions, the effective spring constant is half of the original, resulting in a frequency of 0.380 s-1.
  • #1
Gold3nlily
42
0

Homework Statement


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

Homework Equations


w = sq(k/m)
w = omega = angular frequency
k = spring constant
m = mass

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

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.
 
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  • #2
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
gneill said:
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?

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
Gold3nlily said:
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?).

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
gneill said:
So both forces are to the right, opposite the direction of the displacement.

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

What, then, is the effective spring constant?
 

1. What is the definition of frequency of oscillation?

The frequency of oscillation is the number of complete cycles or vibrations of a system in a given unit of time. In the case of a spring, it refers to the number of back-and-forth movements the spring makes in one second.

2. How is the frequency of oscillation calculated?

The frequency of oscillation is calculated by dividing the number of cycles by the time it takes to complete those cycles. In the case of a spring, the frequency is equal to 1 divided by the period (time for one complete cycle) of the oscillation.

3. What factors affect the frequency of oscillation in a spring?

The frequency of oscillation in a spring is affected by its stiffness (spring constant), mass attached to the spring, and the force applied to the spring. The stiffer the spring, the higher the frequency. The greater the mass or force, the lower the frequency.

4. How does the frequency of oscillation change with changes in the spring's length?

According to Hooke's Law, the frequency of oscillation in a spring is inversely proportional to its length. This means that as the length of the spring increases, the frequency decreases, and vice versa.

5. What is the relationship between the frequency of oscillation and the amplitude of a spring's oscillation?

The frequency of oscillation in a spring is not affected by the amplitude (maximum displacement from equilibrium) of its oscillation. This means that even if the amplitude changes, the frequency remains the same as long as other factors, such as mass and stiffness, remain constant.

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