Question

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

Answer 1

Let `T = kl^a g^b m^c …(i)`
where k is dimensionless constant and a,b,c,
are the respecitve dimensious, Writing the
dimensions on both sides of (i), we get
`[M^0L^0 T^1] = L^a (LT^(-2))^b (M^c)`
` = M^c L^(a +b) T^(-2b)`
Equating the dimensious of M,L and T on both
sides, we get
`c = 0, a +b = 0 and -2b = 1 or b= -(1)/(2)`
`:. a= -b =(1)/(2)`
From (i), `T = kl^(1//2). g ^(-1//2. m^0 =k sqrt((I)/(g))`
By other methods, we find `k =2pi`
`:. T = 2pi sqrt(I//g)`