Question

A light spring with spring constant $$k_{1}$$ is hung from an elevated support. From its lower end a second light spring is hung, which has spring constant $$k_{2}$$. An object of mass $$m$$ is hung at rest from the lower end of the second spring. (a) Find the total extension distance of the pair of springs. (b) Find the effective spring constant of the pair of springs as a system.

Answer 1

(a) The force $$mg$$ is the tension in each of the springs. The bottom of the upper (first) spring moves down by distance $$x_{1} = |F|/k_{1} = mg/k_{1}$$. The top of the second spring moves down
by this distance, and the second spring also stretches by $$x_{2} = mg/k_{2}$$. The bottom of the lower spring then moves down by distance
$$x_{total}=x_{1}+x_{2}=\dfrac{mg}{k_{1}}+\dfrac{mg}{k_{2}}=mg\left ( \dfrac{1}{k_{1}}+\dfrac{1}{k_{2}} \right )$$
(b) From the last equation we have
$$mg=\dfrac{x_{1}+x_{2}}{\dfrac{1}{k_{1}}+\dfrac{1}{k_{2}}}$$
This is of the form
$$|F|=\left ( \dfrac{1}{1/k_{1}+1/k_{2}} \right )(x_{1}+x_{2})$$
The downward displacement is opposite in direction to the upward force the springs exert on the load, so we may write $$F = –k_{eff} x_{total}$$, with the effective spring constant for the pair of springs given by
$$k_{eff}=\dfrac{1}{1/k_{1}+1/k_{2}}$$