Question

Consider a simple pendulum consisting of a massive bob suspended from a fixed point by a string. Let T denote the time (period of the pendulum) that it takes the bob to complete one cycle of oscillation. How does the period of the simple pendulum depend on the quantities that define the pendulum and the quantities that determine the motion? [You need to perform a dimensional analysis to solve this one. Start by assuming T = k Inmpgq, where k is a constant and n, p, q needs to be determined. Then use the following table to translate this equation into a dimensional equation by using appropriate dimensional formulas.] Table: Dimensions of quantities that may describe the period of pendulum 1. frictionless pivot bob's....---'massive bob trajectory bob,- :- equilibrium position Name of Quantity Symbol Dimensional Formula Time/period of the pendulum Length of the pendulum Mass of the pendulum Gravitational Acceleration LT-2

Answer 1

T = Kl^n m^p g^2 T M^0 L^0 T^1 l = M^0 L^1 T^0 m = M^1 L^0 T^0 g = M^0 L^1 T^-2 T = Kl^n m^p g^q [M^0 L^0 T^1] = K [M^0 L^1 T^0]^n [M^1 L^0 T^0]^p [M^0 L^1 T^-2]^q [M^0 L^0 T^1] = K [M^p L^n + q T^-2q] On comparing on both sides p = 0 n + q = 0 -2q = 1 q = -1/2 p = 0 n = 1/2 T = K l^1/2 g^-1/2 T = K squareroot l/g