Deriving expression for k_eff

[ksmith159]

See attached file:
I would like to show theoretically that the first double spring system is equivalent to the second system with only one spring keff. Then I want to derive an expression for keff in terms of k1 and k2.

I think this system is in series and therefore keff =1/k1 +1/k2
but that is probably not even close to the right track...

 

[rock.freak667]

Well for the new spring system

$$F=k_{eff}(x_1+x_2)$$

knowing that F is the same for both springs, can you make an attempt now?

(sorry if I sound harsh,am a bit handicapped at the moment and typing is tedious)

 

[kerimek]

First, let's define the two systems in question. The first system consists of two springs (k1 and k2) connected in series, with one end of each spring attached to a fixed point and the other end attached to a mass. The second system consists of only one spring (keff) attached to the same fixed point and mass. Our goal is to show that these two systems are equivalent, meaning they produce the same displacement for the same applied force.

To derive an expression for keff, we can use the concept of equivalent stiffness. This means that the total stiffness of the two springs in the first system is equal to the stiffness of the single spring in the second system. Mathematically, this can be represented as:

k1 + k2 = keff

Now, let's consider the displacement of the mass in each system. In the first system, the displacement is equal to the sum of the displacements of each spring, which can be represented as:

x = x1 + x2

Where x1 and x2 are the displacements of k1 and k2, respectively. In the second system, the displacement is simply equal to the displacement of the single spring, which can be represented as:

x = xeff

Since the displacements in both systems are equal, we can equate these two expressions:

x1 + x2 = xeff

Now, we can use Hooke's Law to relate the displacements to the applied force. Hooke's Law states that the force applied to a spring is directly proportional to its displacement and its stiffness. Mathematically, this can be represented as:

F = kx

Substituting this into our previous equations, we get:

k1x1 + k2x2 = kxeff

Since we know that k1 + k2 = keff, we can substitute this into the equation above:

k1x1 + k2x2 = keffxeff

Now, we can rearrange this equation to solve for keff:

keff = (k1x1 + k2x2)/xeff

We can also express the displacements in terms of the applied force using Hooke's Law:

x1 = F/k1 and x2 = F/k2

Substituting these into our equation for keff, we get:

keff = (k1F/k1 + k2F/k2)/(F/keff)