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?

Answer 1

\[k _{1}x _{1} =k _{2}x _{2} = F\]
\[\frac{k _{2}x _{2} }{ k _{1}} =x _{1}\]
\[F = k _{eq}x = k _{eq}(x _{1} +x _{2})\]

sub.s in for x1:
\[F = k _{eq}(\frac{ k _{2} x _{2} }{ k _{1}} +x _{2})\]

sub.s in for F (in terms of k2 and x2)
\[k _{2}x _{2}= k _{eq}(\frac{ k _{2} x _{2} }{ k _{1}} +x _{2})\]

x2 is in every term:
\[k _{2}= k _{eq}(\frac{ k _{2} }{ k _{1}} +1)\]

multiply both sides by
\[(\frac{ k _{1}}{k _{2} + k _{1} } )\]
\[k _{eq} = (\frac{ k _{1} }{k _{1} + k _{2}})k _{2}\]
\[k_{eq}=(\frac{ 1}{k _{1}} +\frac{ 1}{k _{2}})^{-1}\]