Question

The potential energy function for a particle executing linear simple harmonic motion is given by $U{(}_x)=\frac{1}{2}k{x}^2$, where $k$ is the force constant. For $k=0.5N{m}^{-1}$,the graph of $U\left(x\right)$ versus $x$ is shown in figure. Show that a particle of total energy $1J$ moving under this potential ${}^{\prime }turnsbac{k}^{\prime }$ when it reaches $x=\pm 2m$.

Answer 1

At extreme oh SHM, KE=0

Total energy =potential energy $=\frac{1}{2}k{x}^2$

$1=\frac{1}{2}\left(0.5\right)x^2$

$4=x^2$

$x=\pm 2$