Time taken by the bob to move from A to C is $t_1$ ">

Figure shows an oscillating pendulum.

Time taken by the bob to move from A to C is $t_1$


Time taken from A to C $=t_{AC}=t_1$

Time taken from C to O $=t_{CO}=t_2$

So, time taken from A to O $=t_{AO}=t_1+t_2$

Similarly, time taken from O to B $=t_{OB}=t_1+t_2$

Total time taken from A to B $=t_{AB}=t_{AO}+t_{OB}=t_1+t_2+t_1+t_2$

$=2(t_1+t_2)$

Similarly, time taken from B to A $=2(t_1+t_2)$

We know that the time period is the time taken from A to B and then B to A.

So, the time period of simple pendulam $T=t{AB}+t_{BA}=2(t_1+t_2)+2(t_1+t_2)$

$=4(t_1+t_2)$

So, option $(d)$ is correct.

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Updated on: 10-Oct-2022

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