A particle travels in simple harmonic motion described by the equation x = 2.0 (50 t + ⁻¹ 0.75) , where x is in centimetres and t is in seconds. The motion is initiated at t = 0 . At what time does the particle come to rest for the second time?

Step-by-step solution

The given equation of simple harmonic motion is x = 2.0 (50 t + ) , where = ⁻¹(0.75) . The particle comes to rest when its velocity is zero: v = dx dt = -100 (50 t + ) = 0 This implies (50 t + ) = 0 , which gives 50 t + = n for n = 1, 2, 3, For the particle to come to rest for the second time, we set n = 2 : 50 t + = 2 t = 2 - 50 = 2 50 - 50 Since = 0.75 = 3 4 , the angle 37^ . Converting to radians, 37 180 0.205 . Substituting this value: t 2 - 0.205 50 = 1.795 50 = 0.0359 s Rounding to two significant figures, we