1.

        Note:  I showed this one in class! Here it is again in Maple.

>	equ1:=c1*sin(omega*t)+c2*cos(omega*t);

>	first_derivative:=diff(equ1,t);second_derivative:=diff(first_derivative,t);

Notice:   second_derivative = (equ1) and so it follows that  satisfies the given differential equation.

2.

          Note:  This I also did in class. Here it is again in Maple.

>	equ2:=M*cos(omega*t+delta);

>	expand(equ2,t);

Just let  and  and the above becomes equ1. (Remember,  is a constant so that  and  are both constants.)

3.

        This one is a straight forward calculation and algebraic re-arrangement. Do it by hand!! Here it is in Maple Syrup:

>	position:=M*cos(omega*t+delta);velocity:=diff(position,t);acceleration:=diff(velocity,t);

>	acceleration/position;

Notice:   acceleration =  so acceleration is directly proportional to displacement.

4.

     This was done in class (a long  time ago -- before calling it the "integrative activity") using calculations. Another way to think about it, without  calculations, is as follows. We know the velocity and acceleration curves are ordinary sine and cosine curves respectively. Therefore, we know that these have their maximum and minimum values only where they have horizontal tangent lines  (this is not true in general -- can you think of reasons why?). Since speed is the absolute value of velocity, we know the speed is at a maximum when the velocity curve is at a max or a min. These are precisely where the tangent lines to its graph are horizontal. The tangent lines are horizontal exactly where the derivative of the velocity function is zero and the derivative of the velocity function is the acceleration function! That is, the speed is at a max when the acceleration is zero. Conversely, when the acceleration is zero, the velocity curve has a horizontal tangent line and this only happens when the velocity is at a maximum or minimum value (since the velocity curve is a sine curve).

5.

        a)

>	position:=1/5*cos(8*t);

>	velocity:=diff(position,t);acceleration:=diff(velocity,t);

>	acceleration/position;

Note:  aceleration = -64(position) so we have that x'' =  where  = -64  ( so that ) .  i.e. the position equation obeys the differential equation  for Hooke's Law.

      b)         Displacement s  is at a maximum from equilibrium when the velocity is zero: i.e. when  and    when . This means the first two time are  and  seconds. The plot below of position vs. time confirms this.

>	plot(position,t=0..2,title=`position vs time`,thickness=2);

               The Speed (recall:  ) is at a maximum when the acceleration is zero (which is where the tangent line to the velocity graph is horizontal). This happens when   which has the solution  or  so that the first two times are  and  and these are confirmeds in the plot below of velocity vs. time.

>	plot(velocity,t=0..2,title=`Velocity vs time`,thickness=2,colour=blue);

6.

>	s:=sin(Pi/2*t)+cos(Pi/2*t); v:=diff(s,t); a:=diff(v,t);