 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-b/8/c/solution

\overset{x}{˙}

is the speed along the track, and

y = B x^{2}

supplies the gravitational potential. The conserved energy is

E = \frac{M}{2} (1 + \frac{r _{2}^{2}}{2 r _{1}^{2}}) \overset{x}{˙}^{2} + M g B x^{2} .

Differentiating and extending through turning points by continuity gives

(1 + \frac{r _{2}^{2}}{2 r _{1}^{2}}) \overset{x}{¨} + 2 g B x = 0.

ω_{0}^{2} = 4 g B / {(r_{2} / r_{1})^{2} + 2}

. Hence

f_{0} = \frac{g B}{π ( r _{2} / r _{1} ) ^{2} + 2} = 0.235 Hz .

The frequency is lower than for a sliding point mass because energy is also stored in rotation.

 New to topics? Read the docs here!