Source: Extra Practice

The period (TT) of oscillation of a simple pendulum depends on its length (ll), mass of the bob (mm), and acceleration due to gravity (gg). Using dimensional analysis, which of the following expressions is dimensionally consistent for the period?

Options

Option A

T∝mlgT \propto m l g

Option B

T∝l/gT \propto l/g

Option C is correct

T∝l/gT \propto \sqrt{l/g}

Option D

T∝ml/gT \propto \sqrt{m l/g}

Explanation

Let's assume T=k⋅lambgcT = k \cdot l^a m^b g^c, where kk is a dimensionless constant. Writing the dimensions for each quantity: [T]=[T1][T] = [T^1] (Period) [l]=[L1][l] = [L^1] (Length) [m]=[M1][m] = [M^1] (Mass) [g]=[LT−2][g] = [LT^{-2}] (Acceleration due to gravity) Substituting these into the assumed relation: [M0L0T1]=[L]a[M]b[LT−2]c[M^0L^0T^1] = [L]^a [M]^b [LT^{-2}]^c [M0L0T1]=[LaMbLcT−2c][M^0L^0T^1] = [L^a M^b L^c T^{-2c}] [M0L0T1]=[MbLa+cT−2c][M^0L^0T^1] = [M^b L^{a+c} T^{-2c}] Comparing the powers of M, L, and T on both sides: For M: b=0b = 0 For L: a+c=0  ⟹  a=−ca+c = 0 \implies a = -c For T: 1=−2c  ⟹  c=−1/21 = -2c \implies c = -1/2 Substitute c=−1/2c = -1/2 into a=−ca = -c: a=−(−1/2)=1/2a = -(-1/2) = 1/2. So, T∝l1/2m0g−1/2=l1/2g−1/2=lgT \propto l^{1/2} m^0 g^{-1/2} = l^{1/2} g^{-1/2} = \sqrt{\frac{l}{g}}.