A particle executes simple harmonic motion with an amplitude of 10 cm and a time period of 6 s . At t = 0 , it is located at x = 5 cm and is moving towards the positive x-direction. Determine the equation for the displacement x as a function of time t , and calculate the magnitude of the particle's acceleration at t = 4 s .

Step-by-step solution

The angular frequency of the particle is = 2 T = 2 6 = 3 rad/s . The general equation for displacement in SHM is x(t) = A ( t + ₀) . Given A = 10 cm , and at t = 0 , x = 5 cm : 10 ( ₀) = 5 ( ₀) = 1 2 ₀ = 6 or 5 6 . The velocity is v(t) = A ( t + ₀) . Since the particle is moving towards the positive x-direction at t = 0 , v(0) > 0 , which implies ( ₀) > 0 . Therefore, ₀ = 6 . The equation for displacement is x(t) = 10 ( 3 t + 6 ) cm . At t = 4 s , the displacement is: x(4) = 10 ( 4 3 + 6 ) = 10 ( 9 6 ) = 10 ( 3 2 )