Question

Two springs with spring constants k1 and k2 are connected together. With mass m1 attached, the springs have
lengths d1 and d2. Increasing the pull mass to m2 stretches the
springs a distance x1 and x2 for a total stretch of x. Derive an
equation for the effective force constant ke of the two spring in
series. (Hint: The stretching force F is the same for both springs.)
What is ke when k1 = k2? Does the k of a spring depend on the
length of the coiled spring?

How do I derive this equation?
And ke is dependent on mass right? not length?

Answer 1

ke is both dependent on Force and length.
First, you'll have to use hooke's law. \(F=kx\)
in this case, \(x=x_1+x_2\)
dividing throughout by F, \(\frac{x}{F}=\frac{x_1}{F}+\frac{x_2}{F} \)
which is equal to, \(\frac{1}{k_e}=\frac{1}{k_1}+\frac{1}{k_2} \)
Do you require further help?

Answer 2

thank you!! :)
I figured out a different way as well.

Answer 3

You're welcome :)
It's similar to the resistance is ohmic conductors.