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Standard XII

Physics

Intro to Simple Pendulum

A thin rectan...

Question

A thin rectangular magnet suspended freely has a period of oscillation $4 s$ . If it is broken into two halves each having half their initial length, then when suspended similarly, the time period of oscillation of each part will be

4 s

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2 s

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1 s

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4 \sqrt{2} s

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Solution

The correct option is D $4 \sqrt{2} s$
A thin rectangular magnet suspended freely has a period of oscillation $T = 4 s$
We know that
$T = 2 π \sqrt{\frac{l}{g}} \frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}} \frac{4}{T_{2}} = \sqrt{\frac{l_{1}}{l_{1} / 2}} = \sqrt{2} o r T_{2} = 4 \sqrt{2} s$

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A thin rectangular magnet suspended freely has a period of oscillation 4 s. If it is broken into two halves each having half their initial length, then when suspended similarly, the time period of oscillation of each part will be

The correct option is D 4√2 sA thin rectangular magnet suspended freely has a period of oscillation T=4sWe know thatT=2π√lgT1T2=√l1l24T2=√l1l1/2=√2orT2=4√2s

A thin rectangular magnet suspended freely has a period of oscillation $4 s$ . If it is broken into two halves each having half their initial length, then when suspended similarly, the time period of oscillation of each part will be

The correct option is D $4 \sqrt{2} s$
A thin rectangular magnet suspended freely has a period of oscillation $T = 4 s$
We know that
$T = 2 π \sqrt{\frac{l}{g}} \frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}} \frac{4}{T_{2}} = \sqrt{\frac{l_{1}}{l_{1} / 2}} = \sqrt{2} o r T_{2} = 4 \sqrt{2} s$