Derive an expression for the time period (T) of a simple pendulum which may depend upon the mass (m) of the bob, length (l) of the pendulum and acceleration due to gravity (g).

Explanation

Let `T= km^(a) l^(b) g^(c )` where k is a dimensionless constant. Writing the equation in dimensional form, we have `[M^(0) L^(0) T^(1)]= [M]^(a) [L]^(b) [LT^(-2)]^(c )= [M^(a) L^(b+ c)T^(-2c)]` Equating exponents of M, L and T on both sides, we get a= 0, b+ c= 0, `-2c= 1`. Solving the eq. we get a= 0, b=1/2, c= `-1//2`
Hence, `T= K sqrt((l)/(g))`, where k is constant.