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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?
```
3 years ago

Jitender Pal
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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