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

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we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .