A particle of mass 5 g is executing S.H.M. with an amplitude 0.3 m and period 5 π s . The maximum value of the force acting on the particle is

m = 5 g = 5 × 1 0 − 3 kg , A = 0.3 m , T = 5 π s ω = T 2 π = 10 rad / s Maximum force, F = m ω 2 A = 5 × 1 0 − 3 × ( 10 ) 2 × 0.3 = 0.15 N

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PhysicsSHMMiscellaneousEasy2 minPYQ_2021

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A particle of mass $5 g$ is executing S.H.M. with an amplitude $0.3 m$ and period $\frac{π}{5} s$ . The maximum value of the force acting on the particle is

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Solution:

$m = 5 g = 5 \times 1 0^{- 3} kg, A = 0.3 m, T = \frac{π}{5} s ω = \frac{2 π}{T} = 10 rad / s$ Maximum force, $F = m ω^{2} A = 5 \times 1 0^{- 3} \times (10)^{2} \times 0.3 = 0.15 N$

Stream:MHT_CETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2021

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