Using dimensional analysis, find the correct relation for time period $T$ of a simple pendulum:
A$T \propto \sqrt{\frac{L}{g}}$
B$T \propto \sqrt{\frac{m}{g}}$
C$T \propto \sqrt{g}\, l$
D$T \propto \frac{L}{g}$
Answer & Solution
Correct answer: A. $T \propto \sqrt{\frac{L}{g}}$
Assume $T\propto L^a g^b$. Then $[T]=[L]^a[LT^{-2}]^b=[L^{a+b}T^{-2b}]$. Matching powers gives $-2b=1\Rightarrow b=-\frac12$, and $a+b=0\Rightarrow a=\frac12$. So $T\propto \sqrt{L/g}$.
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