A harmonic oscillator is described by the function x(t) 5 (0.200 m) cos (0.350t). Find the oscillator’s (a) maximum velocity and (b) maximum acceleration. Find the oscillator’s (c) position, (d) velocity, and (e) acceleration when t 5 2.00 s

a) Maximum velocity = 0.07 m/s

b) Maximum acceleration = 0.0245 m/s²

c) Position at t = 2s is 0.153 m

d) velocity at t = 2s is - 0.0451 m/s

e) Acceleration at t = 2s is 0.0187 m/s²

Explanation:

x(t) = (0.200 m) cos (0.350t)

The general expression for the displacement of a body in simple harmonic motion = A cos wt

Comparing, A = amplitude = maximum displacement = 0.20 m

w = angular frequency/velocity = 0.350 rad/s

a) v = dx/dt = (d/dt) (A cos wt) = - Aw sin wt

Maximum velocity is given by Aw = 0.2 × 0.35 = 0.07 m/s

b) a = dv/dt = (d/dt) (-Aw sin wt) = - Aw² cos wt

Maximum acceleration = Aw² = 0.2 × 0.35² = 0.0245 m/s²

c) Position at t = 2,

x = A cos wt = 0.2 cos (0.35t)

At t = 2

x = 0.2 cos (0.35 × 2) = 0.2 cos (0.7) = 0.153 m (Take note, wt is in radians)

d) velocity at t=2s

v = - Aw sin wt = - (0.2 × 0.35) sin (0.35 × 2) = - 0.07 sin (0.7) = - 0.0451 m/s

e) Acceleration at t = 2s

a = - Aw² cos wt = - (0.2 × 0.35²) cos (0.35 × 2) = - 0.0245 cos (0.7) = - 0.0187 m/s²

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.