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 acceleration attain its maximum magnitude for the first time?

Step-by-step solution

The displacement of the particle is given by x = 2.0 (50 t + ⁻¹ 0.75) . The acceleration of a particle in simple harmonic motion is a = - ^2 x . The magnitude of acceleration is maximum when the displacement is at its extreme positions, i.e., x = 2.0 . This condition is satisfied when (50 t + ⁻¹ 0.75) = 1 . Let = ⁻¹(0.75) = ⁻¹ ( 3 4 ) . The angle is approximately 37^ or 37 180 radians. At t = 0 , the phase is 37^ . The first time the magnitude of acceleration reaches its maximum is when the phase equals (or 180^ ).