Using dimensional analysis, derive the time period T of a simple pendulum in terms of the bob mass m, pendulum length l, and gravitational acceleration g.

What dimensional analysis can and cannot do here

Dimensional analysis can tell you how $T$ scales with $m$, $l$, and $g$. It cannot give the exact numerical constant (like $2\pi$); it only gives a dimensionless factor $k$.

Assume a power-law dependence

Let the time period depend on $m$, $l$, and $g$ as $$T = k\, m^a l^b g^c,$$ where $k$ is a dimensionless constant and $a, b, c$ are powers to be found.

Write dimensions of each quantity

• $[T] = T$
• $[m] = M$
• $[l] = L$
• $[g] = LT^{-2}$

So the right-hand side has dimensions: $$[m^a l^b g^c] = M^a L^b (LT^{-2})^c = M^a L^{b+c} T^{-2c}.$$

Match exponents of $M$, $L$, and $T$

Equate dimensions of both sides: $$T^1 = M^a L^{b+c} T^{-2c}.$$ This gives three equations by matching powers:

• For $M$: $a = 0$
• For $T$: $-2c = 1 \Rightarrow c = -\tfrac12$
• For $L$: $b + c = 0 \Rightarrow b = \tfrac12$

Final result

Substitute $a=0$, $b=\tfrac12$, $c=-\tfrac12$ into $T = k m^a l^b g^c$: $$T = k\, l^{1/2} g^{-1/2} = k\sqrt{\frac{l}{g}}.$$ So the period does not depend on the bob mass $m$ in this model.