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Math Help - lightly damped oscillators

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    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!
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    Quote Originally Posted by macabre View Post
    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
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