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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?

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?

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
9 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.

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