SHMHard
Question
The displacement (in m) of a particle of mass 100 g from its equilibrium position is given by the equation:
y = 0.05 sin 3π (5t + 0.4)
y = 0.05 sin 3π (5t + 0.4)
Options
A.the time period of motion is 
sec
secB.the time period of motion is 
sec.
sec.C.the maximum acceleration of the particle is 11.25π2 m/s2
D.the force acting on the particle is zero when the displacement is 0.05 m.
Solution
ω = 3π × 5 = 15π, A = 0.05
T =
sec. Ans. ---------(B)
amax = ω2 A
= (15π)2 0.05
= 225 × 0.05
= 11.25 π2 Ans.---------(C)
T =
sec. Ans. ---------(B)amax = ω2 A
= (15π)2 0.05
= 225 × 0.05
= 11.25 π2 Ans.---------(C)
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