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PhysicsSHMCalculation of time periodMedium2 minQB

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$m_{1}$ & $m_{2}$ are connected with a light inextensible string with $m_{1}$ lying on smooth table and $m_{2}$ hanging as shown in figure. $m_{1}$ is also connected to a light spring which is initially unstretched and the system is released from rest :-

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Answer:

Solution:

At the new equilibrium, $m_{2} g = k x_{0}$ . For a small extra displacement x, the restoring force kx acts on total mass $m_{1} + m_{2}$ :

$a = \frac{k x}{m _{1} + m _{2}} = ω^{2} x \Rightarrow ω = \frac{k}{m _{1} + m _{2}}$ - SHM. Hence (B).

(Tension is not zero at release, and maximum displacement of $m_{1}$ is $2 x_{0} = \frac{2 m _{2} g}{k}$ , so (C) and (D) are wrong.)

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: QB

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