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Standard XII

Physics

Springs in Parallel and Series

A 1 kg weig...

Question

A $1 k g$ weight is suspended to a mass-less spring and it has a period $T$ . If now a $4 k g$ is suspended from the same spring the new period will be

4 T

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2 T

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T

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T / 2

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Solution

The correct option is B $2 T$

$T = 2 π \sqrt{\frac{m}{k}}$
$T_{1} = T$
$m_{1} = 1 k g$
$m_{2} = 4 k g$
$T = 2 π \sqrt{\frac{1}{k}}$
$T_{2} = 2 π \sqrt{\frac{4}{k}}$
$T_{2} = 2 T$

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A 1 kg weight is suspended to a mass-less spring and it has a period T. If now a 4 kg is suspended from the same spring the new period will be

The correct option is B 2T T=2π√mkT1=Tm1=1kgm2=4kgT=2π√1kT2=2π√4kT2=2T

A $1 k g$ weight is suspended to a mass-less spring and it has a period $T$ . If now a $4 k g$ is suspended from the same spring the new period will be

The correct option is B $2 T$

$T = 2 π \sqrt{\frac{m}{k}}$
$T_{1} = T$
$m_{1} = 1 k g$
$m_{2} = 4 k g$
$T = 2 π \sqrt{\frac{1}{k}}$
$T_{2} = 2 π \sqrt{\frac{4}{k}}$
$T_{2} = 2 T$