1.A simple pendulum performs simple harmonic motion about x=0x = 0x=0 with an amplitude aaa and time period TTT. The speed of the pendulum at x=a/2x = a/2x=a/2 will bea.πa3/T\pi a \sqrt{3} / Tπa3/Tb.πa3/(2T)\pi a \sqrt{3} / (2T)πa3/(2T)c.πa/T\pi a / Tπa/Td.3π2a/T23\pi^2 a / T^23π2a/T2Login to continueOnly logged in users canattempt or see the solution.
