Two simple pendulums A and B have lengths 1.0 m and 4.0 m respectively at a certain place. Which pendulum will make more oscillations in 1 minute? Explain your answer.

Given,

l_A = 1.0m
l_B = 4.0m

Since,

$T \propto \sqrt{l} \\[0.5em]$

Therefore,

$\dfrac {T$A}{TB} = \dfrac{\sqrt{lA}}{\sqrt{lB}} \\[0.5em]

Substituting the values of l in the formula above we get,

So,

$\dfrac {T$A}{TB} = \dfrac{\sqrt{1}}{\sqrt{4}} \\[0.5em] \Rightarrow \dfrac {TA}{TB} = \dfrac{1}{2} \\[0.5em]

i.e., T₁ : T₂ = 1 : 2

Therefore, time period of B is more (twice) than that of A. Hence, A will make more oscillations (twice) in a given time than B.