Physics - SHM Question with Solution | TestHub

PhysicsSHMAngular SHMEasy2 minQB
PhysicsEasynumerical

A horizontal rod of massm=3kπ2 kgand lengthLis pivoted at one end. The rod at the other end is supported by a spring of force constantk. The rod is displaced by a small angleθfrom its horizontal equilibrium position and released. The time period (in second) of the subsequent simple harmonic motion is


Answer:
2.00
Solution:

E=12mL23ω2+12kx-x02+mgx2

dEdt=12mL232vLaL+kx-x0v+mg2v=0

=m3a+kx-kx0+mgv=0

At equilibrium,

kx0=mg2

∴ma3=-kx

a=-3kmx

T=2πω=2πm3k

=2π3k3kπ2=2 s

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Angular SHM
⏱ 2mℹ️ Source: QB

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