Question

When the displacement of a mass on a spring is one half of the amplitude of its oscillation, what fraction of the total energy is potential energy?

Answer 1

Question #17210

For a mass on a spring, $x = A\cos (\omega t + \varphi), \omega = \sqrt{\frac{k}{m}}$ . Total energy includes kinetic and potential energy: $E = K + U = \frac{mx^2}{2} + \frac{kx^2}{2}$ . Plugging the first expression into this expression, obtain:

$$E = \frac {m A ^ {2} \omega^ {2}}{2}. \text { Hence, for } x = A / 2, \frac {U}{E} | _ {x = A / 2} = \frac {\omega^ {2} m x ^ {2}}{2} \cdot \frac {2}{m A ^ {2} \omega^ {2}} | _ {x = A / 2} = \frac {x ^ {2}}{A ^ {2}} | _ {x = A / 2} = \frac {1}{4}.$$