Hooke's law - addition of force constant

[ys2050]

I need to figure out an equation that shows the relationship between the force constants of two springs that are connected together.
I know that the 1/ktot = 1/k1 + 1/k2
but I have to shows all the steps to obtain this equation...
I know that
F = k1x1 = k2x2 = k(x1 + x2)
I tried manipulating the equation to get the expression I want but i don't seem to be getting it... :(
Please help!

 

[Dick]

That's not 'the' equation. That's two equations. Solve k1*x1=k2*x2 for x2. Put that into k1*x1=k*(x1+x2). Cancel the common factor of x1 and solve for k in terms of k1 and k2.

 

[Dick]

Good point. You are probably asking why, for example, if I hook two springs with spring constant k together, the resulting spring constant is k/2? Because to get to a given displacement, each spring has to stretch only half as far. The material properties didn't change. Imagine two springs made of the same stuff, one 1m long and the other 2m long. It really is easier to stretch the long spring by an extra meter than the short spring.

 

[Dick]

Hey, this is just like a couple of other questions asked about the spring constant of a wire last night. If the bulk modulus of the material is M, then the spring constant of a wire is k=M*A/L, where A is the cross sectional area and L is the length. Double L and you halve k.

 

[ys2050]

Good point. You are probably asking why, for example, if I hook two springs with spring constant k together, the resulting spring constant is k/2? Because to get to a given displacement, each spring has to stretch only half as far. The material properties didn't change. Imagine two springs made of the same stuff, one 1m long and the other 2m long. It really is easier to stretch the long spring by an extra meter than the short spring.

hmmm I'm not sure if I'm getting it...
so in my case, the total spring did not have to exert as much force in order for the spring to get to a certain distant because...??

 

[Dick]

Because the 2m spring has to flex the material less when stretched by one meter than the 1m spring, if you want to put it that way. Your derivation is correct. If you have access to stuff, hook two springs together and measure. You'll see you were right.