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Physics - SHM Question with Solution | TestHub

PhysicsSHMMiscellaneousMedium3 minQB

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If a 2 kg mass is attached to a spring of force constant $k = 1250 N / m$ , the period of oscillation is $T = \frac{π}{12}$ sec. The damping constant 'b' has the value :

Options:

Answer:

Solution:

$ω_{D} = \frac{2 π}{T} = 24 rad/s ω_{0} = \frac{k}{m} = 25 rad/s ω_{D}^{2} = ω_{0}^{2} - γ^{2} γ = 625 - 576 = 7 s^{- 1} b = 2 mγ b = 2 \times 2 \times 7 = 28 kg/s Option (D)$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Miscellaneous

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