Question icon
Grade 10Mechanics

Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos(50 π t + tan-1 0.75) Where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the first time? (b) When does the acceleration have its maximum magnitude for the first time? (c) When does the particle come to rest for the second time?

Profile image of Hrishant Goswami
12 Years agoGrade 10
Answers icon

1 Answer

Profile image of Jitender Pal
12 Years ago
Sol. . a) x = 2.0 cos (50πt + tan–1 0.75) = 2.0 cos (50πt + 0.643) V = dx/dt = - 100 sin (50π + 0.643) ⇒ sin (50πt + 0.643) = 0 As the particle comes to rest for the 1st time ⇒ 50πt + 0.643 = π ⇒ t = 1.6 × 10–2 sec. b) Acceleration a = dv/dt = - 100π × 50 π cos (50πt + 0.643) For maximum acceleration cos (50πt + 0.643) = – 1 cos π (max) (so a is max) ⇒ t = 1.6 × 10–2 sec. c) When the particle comes to rest for second time, 50πt + 0.643 = 2π ⇒ t = 3.6 × 10–2 s.