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)
Create a free account to view solution
View Solution FreeMore SHM Questions
Two particles are oscillating along close parallel straight lines side by side, with the same frequency and amplitudes. ...Five identical springs are used in the following three configurations. The time periods of vertical oscillations in conf...Two simple pendulums, having periods of 2s and 3s respectively, pass through the mean position simultaneously at a parti...A simple pendulum is suspended from the ceiling of a vehicle, its time period is T. Vehicle is moving with constant velo...A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by...