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 $x$:
\begin{eqnarray}
& & 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}}
\end{eqnarray}
Powered by Hirophysics.com
No comments:
Post a Comment