Two simple pendulums first $(A)$ of bob mass $M$ and length $L_{1}$ , second $(B)$ of bob mass $M_{1}$ and length $L_{2}$ . $M_{1} = M_{2}$ and $L_{1} = 2 L_{2}$ . If the vibrational energy of both is same. Then which of the following is correct ?

Amplitude of

B

is greater than that of

A

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Amplitude of

B

is smaller than that of

A

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Amplitude will be same

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None of the above

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Solution

The correct option is B Amplitude of $B$ is smaller than that of $A$
$M_{1} = M_{2} = M$
$T_{A} = 2 π \sqrt{\frac{L_{1}}{g}} = 2 π \sqrt{\frac{2 L_{2}}{g}}$
$T_{B} = 2 π \sqrt{\frac{L_{2}}{g}}$
$T_{A} > T_{B}$ , restoring force being the same for both of them
So, Amplitude of B is smaller than that of A

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