Question

The displacement x with respect to time t of a particle moving in simple harmonic motion is given by x = 5cos(16pi*t). Where x is in mm and t is in seconds. If the particle starts at x = 5mm and t = 0s. At what time t does it first pass through its equilibrium position?

A) 1/32 s

B) 1/16 s

C) 1/5 s

D) 4 s

E) 8 s

Answer 1

The given equation of displacement with respect to time of a particle moving in simple harmonic motion is,x= 5cos(16πt)For the particle to pass through its equilibrium position, the displacement value must be zero. Therefore, we can equate x to zero and solve for t.0 = 5cos(16πt)cos(16πt) = 0Now, cos(θ) = 0 for θ = (2n + 1)π/2, where n is an integer.Therefore, 16πt = (2n + 1)π/2⇒ t = (2n + 1)/32, where n is an integer.Since the particle starts at x = 5 mm and moves in the negative x-direction, it first passes through its equilibrium position when its displacement is zero, which occurs at t = (2n + 1)/32 s.Substituting n = 0, we get t = (2(0) + 1)/32 = 1/32 s.Therefore, the particle first passes through its equilibrium position at t = 1/32 s.Option (A) is the correct answer.