A 300-g object attached to the end of a spring oscillates with an amplitude of 7.0 cm and a frequency of 1.80 Hz. (a) Find its maximum speed and maximum acceleration. (b) What is its speed when it is 3.0 cm from its equilibrium position

a) v = 0.79 m / s,  a = 8.95 m / s², b)     v = -0.3285 m / s

Explanation:

A simple harmonic motion is described by the expression

        x = A cos (wt + Ф)

the range of motion is A = 7.0 cm = 0.070 m

angular velocity and frequency are related

         w = 2πf

         w = 2π 1.80

         w = 11.3 rad / s

we substitute

        x = 0.070 cos (11.3t +Ф)

a) to find the velocity we use

       v = dx / dt

        v = - Aw sin (wt + Ф)

the maximum velocity when the cosine argument is π/2 0 3π/2 therefore the sine function is ±1

         v = A w

         v = 0.070 11.3

         v = 0.79 m / s

the acceleration is

         a = dv / dt

         a = - A w² cos (wt + Ф)

the acceleration is maximum for an angle of o or pi, consequently the cosine works worth ±1

         a = A w²

         a = 0.070 11.3²

         a = 8.95 m / s²

b) Let's find the time it takes to get to x = 0.03 m

        wt + Ф = x / A

         wt + Ф = 0.03 / 0.07 = 0.42857

To find the value of fi the initial conditions are used, in general if the system is released from rest fi = 0

          t = 0.42857 / w

          t = 0.42857 / 11.3

          t = 0.0379 s

the speed for this time is

          v = -A w sin wt

           v = - 0.07 11.3 sin (11.3  0.0379)

remember angles are in radians

           v = -0.3285 m / s

a straight line sloped