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Standard IX

Physics

Simple Pendulum

A simple pend...

Question

A simple pendulum has a period of $T_{1}$ when its length is $L$ and its mass is $M$ . A second simple pendulum has a period of $T_{2}$ when its length is $4 L$ and its mass is $2 M$ .
What is the ratio of the periods, $\frac{T_{2}}{T_{1}}$ ?

\frac{1}{2}

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1

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\sqrt{2}

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2

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2 \sqrt{2}

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Solution

The correct option is E $2$
Given : $M_{1} = M$ $L_{1} = L$ $M_{2} = 2 M$ $L_{2} = 4 L$
Time period of simple pendulum $T = 2 π \sqrt{\frac{L}{g}}$ $⟹$ $T \propto \sqrt{L}$
$∴$ $\frac{T_{2}}{T_{1}} = \sqrt{\frac{L_{2}}{L_{1}}}$

$⟹$ $\frac{T_{2}}{T_{1}} = \sqrt{\frac{4 L}{L}} = 2$

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A simple pendulum has a period of T1 when its length is L and its mass is M. A second simple pendulum has a period of T2 when its length is 4L and its mass is 2M.What is the ratio of the periods, T2T1?

The correct option is E 2Given : M1=M L1=L M2=2M L2=4LTime period of simple pendulum T=2π√Lg ⟹ T∝√L∴ T2T1=√L2L1⟹ T2T1=√4LL=2

A simple pendulum has a period of $T_{1}$ when its length is $L$ and its mass is $M$ . A second simple pendulum has a period of $T_{2}$ when its length is $4 L$ and its mass is $2 M$ .
What is the ratio of the periods, $\frac{T_{2}}{T_{1}}$ ?

The correct option is E $2$
Given : $M_{1} = M$ $L_{1} = L$ $M_{2} = 2 M$ $L_{2} = 4 L$
Time period of simple pendulum $T = 2 π \sqrt{\frac{L}{g}}$ $⟹$ $T \propto \sqrt{L}$
$∴$ $\frac{T_{2}}{T_{1}} = \sqrt{\frac{L_{2}}{L_{1}}}$

$⟹$ $\frac{T_{2}}{T_{1}} = \sqrt{\frac{4 L}{L}} = 2$