## Essential University Physics: Volume 1 (3rd Edition)

a) We simplify the equation for energy to find the maximum angular displacement: $\theta_{max} = \sqrt{\frac{2E}{k}}= \sqrt{\frac{2(4.7)}{3.4}}=\fbox {1.66 rads}$ b) We find the maximum angular speed: $\omega= \sqrt{\frac{k}{I}} = \sqrt{\frac{3.4}{1.6}}=1.46$ This means that the maximum speed is: $\omega\times \theta=1.46 \times 1.66 = \fbox{2.42 radians per second}.$