91Ó°ÊÓ

First, we need to find the length of the pendulum. It is possible to get the length by utilizing the information given: When the pendulum passes through the equilibrium point, it has maximum kinetic energy, and when it is at its maximum displacement (amplitude), it has maximum potential energy. By using the conservation of energy principle, we have: \(E_{k}=E_{p}\) \(\frac{1}{2} m v^{2}=mgh\) where \(m\) is the mass of the pendulum bob, \(v\) is the initial velocity, \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^{2}}\)), and \(h\) is the height of the pendulum bob at its maximum displacement. Note that the mass of the bob (m) cancels out. Since \(A = 0.20 m\) and we assume the amplitude is small, then \(h \approx \frac{1}{2}A\). Therefore, \(\frac{1}{2} v^{2}=g(\frac{1}{2}A)\) Now, we can solve the equation to find the length \(L\) of the pendulum: \(L = \frac{h}{\sin{\theta}} \approx \frac{A}{\theta}\), \(\theta = \sin^{-1}(\frac{2h}{A})\)

To find the period T, we can use the formula for the period of a small-angle oscillation: \(T=2\pi\sqrt{\frac{L}{g}}\) Now, we have all the required information to calculate T.

Substitute the values of velocity and amplitude into the equation for the length of pendulum and then use the length to find the period: \(v=0.50 \mathrm{m/s}\), \(A=0.20\mathrm{m}\) From step 1: \(\frac{1}{2} (0.50 \mathrm{m/s})^{2} = 9.8\mathrm{m/s^{2}}(\frac{1}{2}h)\) The result gives \(h = 0.01276\mathrm{m}\). Now find \(\theta\): \(\theta = \sin^{-1}(\frac{2 \times 0.01276\mathrm{m}}{0.20\mathrm{m}})\) which gives \(\theta \approx 0.1275 \, \text{rad}\). Now we can find the length \(L\): \(L \approx \frac{0.20\mathrm{m}}{0.1275} = 1.569\mathrm{m}\) Finally, calculate the period T using the formula from step 2: \(T=2\pi\sqrt{\frac{1.569\mathrm{m}}{9.8\mathrm{m/s^{2}}}}\) This gives us: \(T\approx 2.53\,\text{s}\) So the period \(T\) of the pendulum is approximately \(2.53\) seconds.