A bead of mass 0.020 kg slides without friction in a valley where its potential energy is U = U₀(1…

(a) For small $x$: $F \approx -\dfrac{U_0}{L}\cdot\dfrac{x}{L} = -\dfrac{U_0}{L^2}x$. So $k = \dfrac{0.050\ \text{J}}{(0.10\ \text{m})^2} = 5.0\ \text{N/m}$.

(b) $\omega = \sqrt{\dfrac{5.0\ \text{N/m}}{0.020\ \text{kg}}} = 15.8\ \text{rad/s}$, and $T = \dfrac{2\pi}{15.8\ \text{rad/s}} = 0.397\ \text{s}$.

(c) The program finds about $0.398\ \text{s}$ at an amplitude of $0.01\ \text{m}$, $0.424\ \text{s}$ at $0.10\ \text{m}$ and $0.528\ \text{s}$ at $0.20\ \text{m}$. For large swings the period grows. The force no longer grows in step with $x$, so the motion is periodic but not simple harmonic.