Combined Spring Constant

[hammadmunawar]

Homework Statement

The mass is able to move in any direction. All springs are preloaded (compressed) to half their allowable loading capacity. Springs are not properly connected to the mass or ground (they are mounted on a rod on which the mass is moving). Ignoring all friction and gravity.
http://[PLAIN]http://i63.tinypic.com/2nl94lx.jpg [Broken]

Homework Equations

See attempt at a solution.

The Attempt at a Solution

Considering x(t) [ displacement ] is positive to the right.

F_right spring 1 = - K x(t) [because it get compressed and pushes to left ]
F_right spring 2 = - K x(t) [because it get compressed and pushes to left ]

F_left spring 1 = K x(t) [because it gets released from preloaded position and pushes to right ]
F_left spring 2 = K x(t) [because it gets released from preloaded position and pushes to right ]

Combined effect seems to be zero which does not seem right.

Request verification please.

Thanks

 

[PeroK]

How did you come up with that formula? Have you tried simplifying it?

 

[hammadmunawar]

i think the two springs on each side are in parallel and the combined spring constant will be (K+K).

then the springs on both sides are in series to each other to the overall effect will be :

1/((1/(K+K))+(1/(K+K)))

 

[PeroK]

i think the two springs on each side are in parallel and the combined spring constant will be (K+K).

then the springs on both sides are in series to each other to the overall effect will be :

1/((1/(K+K))+(1/(K+K)))

Try simplifying it. Hint $K + K = 2K$

 

[hammadmunawar]

yes it can be simplified. The overall answer is K.

but i want to ask if the approach is correct.

 

[PeroK]

yes it can be simplified. The overall answer is K.

but i want to ask if the approach is correct.

Do you believe that? Having to work against four independent springs is no harder than working against one?

The approach is not correct because you did not consider forces.

 

[Let'sthink]

The answer is wrong displace the mass m by x and then compute the net restoring force acting on the displaced mass. Compressed springs will push the mass to equilibrium position and the elongated springs will pull mass to equilibrium position. Think it over!

 

[hammadmunawar]

Ok.

Well in my case the springs are preloaded to half of their allowable length. So. when the mass moves to the right (For example), the springs on the right will be compresses and push the mass to the left each with a force equal to -kx.

However, the springs at the left will now be released from their preloaded condition and push the mass to the right each with a force kx.

The combined force comes to be zero which confuses me.

 

[PeroK]

Ok.

Well in my case the springs are preloaded to half of their allowable length. So. when the mass moves to the right (For example), the springs on the right will be compresses and push the mass to the left each with a force equal to -kx.

However, the springs at the left will now be released from their preloaded condition and push the mass to the right each with a force kx.

The combined force comes to be zero which confuses me.

It doesn't come to 0. You need to be careful with the signs of the forces on either side.

 

[hammadmunawar]

It doesn't come to 0. You need to be careful with the signs of the forces on either side.

Considering x(t) [ displacement ] is positive to the right.

F_right spring 1 = - K x(t) [because it get compressed and pushes to left ]
F_right spring 2 = - K x(t) [because it get compressed and pushes to left ]

F_left spring 1 = K x(t) [because it gets released from preloaded position and pushes to right ]
F_left spring 2 = K x(t) [because it gets released from preloaded position and pushes to right ]

The combined effect seems to be zero.

 

[PeroK]

Considering x(t) [ displacement ] is positive to the right.

F_right spring 1 = - K x(t) [because it get compressed and pushes to left ]
F_right spring 2 = - K x(t) [because it get compressed and pushes to left ]

F_left spring 1 = K x(t) [because it gets released from preloaded position and pushes to right ]
F_left spring 2 = K x(t) [because it gets released from preloaded position and pushes to right ]

The combined effect seems to be zero.

Are the forces on the right increasing or decreasing?

Are the forces on the left increasing or decreasing?

It's better to use $\Delta F$ for the change in the force.

 

[hammadmunawar]

Are the forces on the right increasing or decreasing?

Are the forces on the left increasing or decreasing?

It's better to use $\Delta F$ for the change in the force.

I think the forces on the right are increasing (mass pushes the springs).

Forces on left are decreasing (springs push the mass, as they are preloaded).

 

[PeroK]

I think the forces on the right are increasing (mass pushes the springs).

Forces on left are decreasing (springs push the mass, as they are preloaded).

Yes. Now, if the overall change in force is zero, as you claim, then either both forces are increasing or both are decreasing.

 

[hammadmunawar]

Ok. So, of we apply any force on the mass it will not encounter any resistance ?

 

[PeroK]

Ok. So, of we apply any force on the mass it will not encounter any resistance ?

No. Let me give you the answer then.

I think the forces on the right are increasing (mass pushes the springs).

Forces on left are decreasing (springs push the mass, as they are preloaded).

The springs on the right are increasing their force to the left. The springs on the left are decreasing their force to the right. So, there is a nett increase of force to the left. Mathematically, with the right being the postive direction:

$\Delta F_{R1} = \Delta F_{R2} = -Kx$ (increasing force to the left)

$\Delta F_{L1} = \Delta F_{L2} = -Kx$ (decreasing force to the right)

$\Delta F_{total} = -4Kx$

 

[hammadmunawar]

So overall the mass will encounter a spring force stiffness equivalent to 4 times the stiffness of individual springs. This makes more sense.

Still I am unable to grasp the concept fully. But I will try.

Thanks for your help.

 

[PeroK]

So overall the mass will encounter a spring force stiffness equivalent to 4 times the stiffness of individual springs. This makes more sense.

Still I am unable to grasp the concept fully. But I will try.

Thanks for your help.

Suppose at the beginning all spings are compressed by $x_0$ from the equilibrium point. Each spring is pushing with a force $kx_0$. The overall force is 0.

If you displace the mass $x_0$ to the right, then the two springs on the right are pushing with a force of $2kx_0$ each. And the two springs on the left are at the equilibrium point, so no force there. The overall force would be $4kx_0$.

 

[hammadmunawar]

ok thanks !