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Physics - NLM Question with Solution | TestHub

PhysicsNLMSimple Application of NLMEasy2 minPYQ_2010

PhysicsEasysingle choice

Let $T_{1}$ and $T_{2}$ be the time periods of springs $A$ and $B$ when mass $M$ is suspended from one end of each spring. If both springs are taken in series and the same mass $M$ is suspended from the series combination, the time period is $T$ , then

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Solution:

$T_{1} = 2 π \frac{M}{k _{1}}$ or $k_{1} = \frac{4 π ^{2} M}{T _{1}^{2}}$ and $k_{2} = \frac{4 π ^{2} M}{T _{2}^{2}}$ In series combination, $k_{eff} = \frac{k _{1} k _{2}}{k _{1} + k _{2}} = \frac{4 π ^{2} M}{T _{1}^{2} + T _{2}^{2}}$ $∴ T = 2 π \frac{M}{k _{eff}} = T_{1}^{2} + T_{2}^{2} .$

Stream:AIIMSSubject:PhysicsTopic:NLMSubtopic:Simple Application of NLM

⏱ 2mℹ️ Source: PYQ_2010

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