## Free Length Spring

Consider a free length spring is compressed to 50 mm with a 1 N applied force. If the 55 mm diameter helical compression spring has a 3 mm wire diameter and 15 coils, what is the spring’s free length in mm? Assume the shear modulus of elasticity is 300 N/mm^2.

Hint
The spring force:
$$F=kx$$$where $$k$$ is the spring constant, and $$x$$ is the spring’s displacement. Hint 2 $$k=\frac{d^4G}{8D^3N}$$$
where $$G$$ is the shear modulus of elasticity, $$d$$ is the wire diameter, $$D$$ is the mean spring diameter, and $$N$$ is the number of active coils.
A spring’s deflection and force are related by $$F=kx$$ where the spring rate (spring constant) $$k$$ is given by:
$$k=\frac{d^4G}{8D^3N}$$$where $$G$$ is the shear modulus of elasticity, $$d$$ is the wire diameter, $$D$$ is the mean spring diameter, and $$N$$ is the number of active coils. Combining the two equations: $$\frac{F}{x}=\frac{d^4G}{8D^3N}$$$
Solving for displacement:
$$x=\frac{8(55mm)^3(15)(1N)}{(3mm)^4(300N/mm^2)}=\frac{19,965,000mm^5}{24,300mm^4}=821.6\:mm$$$Recall that displacement is: $$x=l_f-l_o$$$
where $$l_f$$ is the final length, and $$l_o$$ is the original length.

Because the spring is compressed, displacement is negative:
$$-821.6mm=55mm-l_o$$$$$l_o=55mm+821.6mm=876.6\:mm$$$
876.6 mm