One end of a spring of force constant K is fixed to a vertical wall and other to a body of mass m resting on smooth horizontal surface. There is another wall at a distance of 3 x 0 from the body. If all the collisions are elastic and spring is compressed by 2 x 0 and released, then the time period of oscillation is

T = 2 π m k t = T 2 + 2 × T 6 = 5 T 6

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PhysicsSHMCalculation of time periodEasy2 minQB

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One end of a spring of force constant $K$ is fixed to a vertical wall and other to a body of mass $m$ resting on smooth horizontal surface. There is another wall at a distance of $\sqrt{3} x_{0}$ from the body. If all the collisions are elastic and spring is compressed by $2 x_{0}$ and released, then the time period of oscillation is

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Solution:

$T = 2 π \sqrt{\frac{m}{k}}$

$t = \frac{T}{2} + 2 \times \frac{T}{6} = \frac{5 T}{6}$

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

⏱ 2mℹ️ Source: QB

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