Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g).Derive the expression for its time period using method of dimensions.

Consider a simple pendulum, having a bob attached to a string, that os

1.	Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g).Derive the expression for its time period using method of dimensions.
Answer» Solution :`T prop l^(a0 g^(B) m^(c)` where `a,b,c,in Z` `:.T= kl^(a)g^(b) m^(c) ""...(1)` where k= DIMENSIONLESS constant CONSIDERING dimensions on both sides we have `[M^(0)L^(0)T^(1)]=[L^(1)]^(a)[L^(1)T^(-2)]^(b)[M^(1)]^(c)` `=M^(c)XXL^(a)xxL^(b)T^(-2b)` `=M^(c)xxL^(a+b)T^(-2b)` Comparing power of `M,c=0` Comparing power of `L,a+b=0` Comparing power of `T,-2b=1` `:. b=-(1)/(2)` and From `a+b=0` `a-(1)/(2)=0` `:.a=(1)/(2)` `:.` Put `a=(1)/(2),b=-(1)/(2),c=0` in equation (1), `T=Kl^((1)/(2))g^((1)/(2))` `=k((l)/(g))^((1)/(2))` `:.Tk= sqrt((l)/(g))` Value of `k=2pi` by experiment, `:.T=2pi sqrt((l)/(g))`