Question:

A pendulum bob is released from the height of \(h\) to hit a spring that creates the force \(F=-kx-bx^3\) in terms of the displacement. If the pendulum has mass \(m\), find the compression displacement of the spring.

Answer:

Find the potential energy of spring. Since it is a conservative force, we integrate it in terms of displacement.

\[  U = -\int F dx = \int kx+bx^3 dx = \frac{1}{2}kx^2+\frac{1}{4}bx^4  \]

The potential energy of the bob is \(mgh\). This can be transferred into the spring energy, so

\[  mgh =  \frac{1}{2}kx^2+\frac{1}{4}bx^4  \]

Rearrange it to solve for \(x\):

\[ x^4 +\frac{2k}{b}x^2 = \frac{4mgh}{b}  \]

\[  \left(x^2+\frac{k}{b}\right)^2 – \frac{k^2}{b^2} = \frac{4mgh}{b}  \]

\[ x^2 + \frac{k}{b} = \sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}}  \]

\[ x  = \sqrt{\sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}} -\frac{k}{b}}  \]