1. ## lightly damped oscillators

Q - in the solution

x(t) = q.exp-at . sin(Gt-d) describing the transients for a lightly damped oscillator, show that the local maxima and minima of q occur at roots of tan(Gt-d) = G/a.

Then by sketching the graph of tan [theta] deduce that these stationary valus occur at equal time intervals pi/G

i desperately need help with this! i havent a clue where to start!

2. Originally Posted by macabre
Q - in the solution

x(t) = q.exp-at . sin(Gt-d) describing the transients for a lightly damped oscillator, show that the local maxima and minima of q occur at roots of tan(Gt-d) = G/a.

Then by sketching the graph of tan [theta] deduce that these stationary valus occur at equal time intervals pi/G

i desperately need help with this! i havent a clue where to start!
Local maxs and mins occur where the derivative of the function is equal to zero. So:
x(t) = q*sin(Gt-d)*e^(-at)

x'(t) = q*cos(Gt-d)*G*e^(-at) + q*sin(Gt-d)*(-a)*e^(-at)

We wish to find at what times this is zero.

0 = q*cos(Gt-d)*G*e^(-at) + q*sin(Gt-d)*(-a)*e^(-at)

Dividing out the common (non-zero) factors gives us:
0 = G*cos(Gt-d) - a*sin(Gt-d)

or
sin(Gt-d)/cos(Gt-d) = G/a

tan(Gt-d) = G/a

Now, given that tan(Gt-d) = G/a any t' such that Gt' - d = Gt - d + (pi) will produce a local max. (The zeros of the tangent function occur with a period of (pi).) This means that the most general t will have a period of (pi)/G.

-Dan