Answer to Question #174555 in Physics for Fhd

Question

The period of a simple pendulum is given as T = 2n 1* g. Here I is the length of the pendulum and g is the gravitational acceleration. Find the values of x and y.

Answer 1

$$[T]=[2\pi l^xg^y]\\ [T]=[ l]^x[g]^y\\ [T]=[ L]^x[LT^{-2}]^y\\-2y=1\\y=-\frac{1}{2}\\x+y=0\\\\x=\frac{1}{2}$$