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

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Solution

The time period in SHM is given by,
$T = 2 π \sqrt{\frac{L}{g}}$
So $l α T^{2}$

given
$L_{a} = 1 m$
$L_{b} = 4 m$
Let
Time period of $A = T_{a}$
Time period of $B = T_{b}$

Ratio of length
$\frac{L_{a}}{L_{b}} = (\frac{T_{a}}{T_{b}})^{2}$

$\frac{1}{4} = (\frac{T_{a}}{T_{b}})^{2}$

$\frac{1}{2} = \frac{T_{a}}{T_{b}}$

oscillation of pendulum A in one-minute $= \frac{60}{1} = 60$

oscillation of pendulum A in one-minute $= \frac{60}{2} = 30$

answer:- pendulum A have more oscillation

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