Why Does the Restoring Force Use k1x + k2x in Spring Oscillations?

In summary, in position A, the tension in the springs is opposite and the overall restoring force is k1x+k2x. In position B, the tension in the springs is the same and the overall restoring force is k1x. In position C, the tension in the springs is opposite and the overall restoring force is k2x.
  • #1
jsmith613
614
0
attachment.php?attachmentid=142547&d=1334684647.png

Looking at the three diagrams we can see that there are three possible situations

(a) mass is to the right of eqm
(b) mass is at eqm
(c) mass is to the left of eqm
(eqm = equilibrium)

Lets look at position (a)

If we consider the tension in both springs:
the tension in the spring on the left has increased by k1x
the tension in the spring on the right has decreased by k2x

So the overall restoring force SHOULD BE k1x - k2x
(the spring on the left is trying to pull it back to the left and the spring on the right is trying to pull it the right hence the forces act in opposite directions)

BUT according to my book the overall restoring force is k1x + k2x...how?

Any help is greatly appreciated

exact quote:
Book
Consider the mass at some point during motion. Let its displacement from eqm be x at that point. One of the two springs has been extended by x and the other has been shortened by x. So compared with eqm one spring has extra tension k1x and the other string has its tension reduced by k2x.
The spring constants for the individual springs are k1 and k2.
The extra tension from one spring combines with the reduced tension from the other to give a restoring force of k1x + k2x.
The restoring force can be written as: F = -kx
where k = k1 + k2
The - sign indicates it acts towards eqm
 
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  • #2
The compressed spring is not pulling away from the eqm but pushing towards. IE it is under compression.
 
  • #3
Integral said:
The compressed spring is not pulling away from the eqm but pushing towards. IE it is under compression.

oh right
thanks
 
  • #4
Integral said:
The compressed spring is not pulling away from the eqm but pushing towards. IE it is under compression.

why is tension not present here?
(I don't understand the explanation given)?
 

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  • #5
jsmith613 said:
why is tension not present here?
(I don't understand the explanation given)?
I don't understand the choices. There are two forces acting on the hanging mass: gravity and the force from the spring. Of course, a stretched spring exerts a tension. So I don't know the intention of having 'tension' vs 'spring' as separate choices: you could call that force the spring force or the tension force exerted by the spring.

And the given explanation reads like gibberish. Where is this question from?
 
  • #6
Integral said:
The compressed spring is not pulling away from the eqm but pushing towards. IE it is under compression.

Even if both springs are under tension all of the time, the restoring force would still be [itex] k_1x + k_2x [/itex]. There's an increased force to the left, because the left spring pulls harder and a decreased force to the right because the right sprint pulls less hard than in the equilibrium position. This will give the same effect as an increased force to the right.
 
  • #7
A linear spring pulled from equilibrium exerts a force k*x. A linear spring compressed from equilibrium exerts a force k*x. Thus, the force is k*x+k*x.

I'm not sure where the issue lays
 
  • #8
By the way, that attached .png is garbage, no offence.
 

1. What is the equation for spring oscillations?

The equation for spring oscillations is F = -kx, where F represents the restoring force of the spring, k is the spring constant, and x is the displacement from equilibrium.

2. What does F = -kx represent in spring oscillations?

In this equation, F represents the restoring force of the spring, which is the force that brings the spring back to its equilibrium position when it is disturbed. The negative sign indicates that the force is always directed towards the equilibrium point.

3. What is the significance of the spring constant, k, in F = -kx?

The spring constant, k, is a measure of the stiffness of the spring. It determines how much force is needed to stretch or compress the spring by a certain amount, x. A higher spring constant means the spring is stiffer and will require more force to be displaced.

4. How does the displacement, x, affect the restoring force in F = -kx?

The displacement, x, is directly proportional to the restoring force, F. This means that as the displacement increases, the restoring force also increases. In other words, the further the spring is stretched or compressed, the stronger the force pulling it back to equilibrium.

5. How can F = -kx be used to describe the motion of a mass-spring system?

F = -kx is used to describe the motion of a mass-spring system by applying Newton's Second Law, which states that the net force on an object is equal to its mass times its acceleration. By setting the restoring force equal to the mass times its acceleration, we can derive the equation for the oscillatory motion of the mass-spring system.

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