Question

3. A damped harmonic oscillator is driven by an external force of the form mfo sin ot. The equation of motion is therefore x + 2ßx + ω x-fo sin dot. carefully explaining all steps, show that the steady-state solution is given by x(t) A() sin at 8) Find A (a) and δ(w).

Answer 1

Ans:

We equation of motion – x” + 2Bx’ + w₀²x = f₀sinwt

Steady-state solution = x(t) = A sin(wt-φ)

So A > 0 is the amplitude of the driven oscillation, whereas φ is the phase lag of this oscillation (with respect to the phase of the piston oscillation). Because

x' = wAcos(wt- φ)

x” = -w²Asin(wt- φ)

Put this values in equation of motion, we get

-w²Asin(wt- φ) + 2BwAcos(wt- φ) + w₀²Asin(wt- φ) = f₀sinwt

(w₀² – w²)Asin(wt- φ) + 2BwAcos(wt- φ) = f₀sinwt

However,

$$ \cos(\omega t-\varphi)\equiv \cos(\omega t) \cos\varphi + \sin(\omega t) \sin\varphi$$ and $$ \sin(\omega t-\varphi) \equiv \sin(\omega t) \cos \varphi - \cos(\omega t) \sin\varphi$$ ,

so we obtain,

(A(w₀² – w²)cosφ + 2BwAsinφ – f₀) sin(wt) + A((w²-w₀²)sinφ + 2Bwcos φ)cos(wt) = 0

The only way in which the preceding equation can be satisfied at all times is if the (constant) coefficients of cos(wt) and sin(wt) separately equate to zero. In other words, if

A(w₀² – w²)cosφ + 2BwAsinφ – f₀ = 0

(w²-w₀²)sinφ + 2Bwcos φ = 0

These two expressions can be combined to give

A = f₀ / [(w₀² – w²)2 + 4B²w²]^1/2

φ = tan^-1 (-2Bw/ (w²-w₀²))