 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/e/solution

E = M g h_{0} = 0.04905 J

. In the high-quality factor regime,

Q = 2 π E /Δ E

, so the required input per period is

Δ E ≃ \frac{2 π}{100} (0.04905) = 3.08 \times 1 0^{- 3} J .

This replaces the small dissipated energy each cycle, rather than supplying the entire stored energy again.

 New to topics? Read the docs here!