Question

Consider a particle moving in simple harmonic motion according to the equation
x = 2.0 cos(50πt+tan⁻¹0.75)
where x is in centimeter 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?

Answer 1

Given equation,

     x = 2.0 cos(50πt+tan⁻¹0.75)

(a). Differentiating this equation with respect to time,

   v = -2 × 50π Sin(50πt+tan⁻¹0.75)

v = 0,

Therefore, Sin(50πt+tan⁻¹0.75) = 0

50πt + tan⁻¹0.75 = π

On solving, t = 1.6 × 10⁻² seconds.

(b). a = -2 × 50π × 50π × Cos(50πt + tan⁻¹0.75)

Maximum acceleration =  -ω²A

-ω²A = -2 × 50π × 50π × Cos(50πt + tan⁻¹0.75)

Cos(50πt + tan⁻¹0.75) = 1

50πt + tan⁻¹0.75 = π

On solving, t = 1.6 × 10⁻² seconds.

(c). The particle comes to rest at extreme positions in an S.H.M.

Since it comes to rest first time at t = 1.6 × 10⁻² s, the second time it will come to rest will be at another extreme which will occur at T/2 time after the first rest.

So the required time = 1.6 × 10⁻² + T/2

= 1.6 × 10⁻²+ 2π/2⍵

= 1.6 × 10⁻² + 2π/(2 × 50π)      

= 3.6 × 10⁻² s

Hope it helps.

Answer 2

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