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Hrishant Goswami Grade: 10
        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?
3 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
										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.
3 years ago
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