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 first time?

Step-by-step solution

The position of the particle is given by: x = 2.0 (50 t + ⁻¹ 0.75) The velocity of the particle is the rate of change of position: v = dx dt = -100 (50 t + ⁻¹ 0.75) The particle comes to rest when its velocity is zero, i.e., v = 0 : (50 t + ⁻¹ 0.75) = 0 For the particle to come to rest for the first time after t = 0 , the phase must be equal to (since the initial phase ⁻¹ 0.75 is between 0 and /2 ): 50 t + ⁻¹ 0.75 = We know that ⁻¹ 0.75 = ⁻¹ ( 3 4 ) 37^ . Converting 37^ to radians: ⁻¹ 0.75 37 180 Substituting this