Well Winching Conundrum

The period of a simple pendulum is $2\pi\sqrt{\frac{L}{g}}$ seconds (where $g=9.8$ ). The frequency of our pendulum is thus $\frac{1}{2\pi\sqrt{\frac{L}{9.8}}}$ . In order to find the total oscillations of such a pendulum in a minute, we need to find the area under the graph of frequency against time between 0 and 60 seconds (since oscillations = frequency × time). However, length ( $L$ ) is not constant, so more work needs to be done. $L$ varies from 20 to 2 metres over 60 seconds, so we can express $L$ in terms of $t$ : $L=20-0.3t$ . Substituting this expression for $L$ , we can integrate the new expression for the frequency of the pendulum from 0 to 60 seconds to find the area under the graph, and thus the total oscillations: $\int_0^{60} \! \frac{1}{2\pi\sqrt{\frac{20-0.3t}{9.8}}} \, \mathrm{d}t$ , which can be rearranged to $\frac{\sqrt{9.8}}{2\pi}\int_0^{60} \! (20-0.3t)^{-0.5}\, \mathrm{d}t$ . Using the chain rule, this can be written as $[2\frac{\sqrt{9.8}}{2\pi}*\frac{10}{3}*(20-03t)^{\frac{1}{2}}]_0^{60}$ , which works out as $≈10.1$ oscillations, so our answer is $\color{#20A900}{\boxed{10}}$ .