Question

The position of a simple harmoic oscillator is given by X = (0.4m) cos (0.5t + 6.28). "t" is measured in seconds. Answer the following questions.

a. what is the period of the oscillation

b. find the maximum velocity and the maximum acceleration of the oscillation

c. find the velocity of the oscillator at t = 4s

d. find the acceleration of the oscillator at t = 4s

Answer 1

Given that

X = (0.4m) cos (0.5t + 6.28)

This similar to the equation

X =Acos(wt+@)

Now comapring both the equations we get

A =0.4m

w =0.5rad/s

a)

The period of oscillations are

w =2pi/T

then T =2pi/w =6.28/0.5 =12.56s

b)

The  maximum velocity and the maximum acceleration of the oscillation is given y

X = (0.4m) cos (0.5t + 6.28).

v =dX/dt =-(0.4m)(0.5)sin(0.5t + 6.28).

then vmax =-Aw =-(0.4)(0.5) =0.2m/s

Then maximum acceleration is given by

amax =dv/dt =-Aw2 =(0.4)(0.5)² =0.1m/s²

c)

find the velocity of the oscillator at t = 4s is given by

v =dX/dt =-(0.4m)(0.5)sin(0.5t + 6.28).

= -(0.2)sin(0.5*4+6.28)=-0.028m/s

d)

find the acceleration of the oscillator at t = 4s is given by

a= dv/dt =-(0.4m)(0.5)²cos(0.5t + 6.28).

=-(0.1) cos(0.5*4+6.28) =0.0989m/s2