Forced Oscillations — Question 8

Section 31-6 · Physics pages 5086–5087

Question 8 The energy stored each cycle returns to its initial value in steady forcing A damped driven circuit obeys L˙i + Ri + q/C = vs(t) with i = ˙q. Its steady state is periodic with the source period T. ∼ vs(t) R C L q i = ˙q Schematic; labels, not drawing lengths, specify the data. (a) Derive an exact instantaneous energy balance. (b) Integrate over a cycle and find the work supplied when i = Ip cos(ωt −ϕ).
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Question 8 - Solution The energy stored each cycle returns to its initial value in steady forcing ∼ vs(t) R C L q i = ˙q (a) Method and calculation i(L˙i + q/C) = vsi −Ri2. ˙U = vsi −Ri2, U = Li2/2 + q2/(2C) . (b) Interpretation and further calculation ∆UT = 0 ⇒Ws,T = ∫T 0 Ri2dt. Ws,T = RI2 pT/2 . Physical check The source replaces dissipation each cycle; it does not continually increase stored energy after transients vanish.

Question 8

The energy stored each cycle returns to its initial value in steady forcing

A damped driven circuit obeys Li̇+Ri+q/C=vs(t)L\dot i+Ri+q/C=v_s(t) with i=q̇i=\dot q. Its steady state is periodic with the source period T.

Diagram shown in the original SVG worksheet.

  1. Derive an exact instantaneous energy balance.

  2. Integrate over a cycle and find the work supplied when i=Ipcos⁡(ωt−ϕ)i=I_p\cos(\omega t-\phi).

Question 8 - Solution

The energy stored each cycle returns to its initial value in steady forcing

Diagram shown in the original SVG worksheet.

(a) Method and calculation

i(Li̇+q/C)=vsi−Ri2.i(L\dot i+q/C)=v_si-Ri^2.U̇=vsi−Ri2,U=Li2/2+q2/(2C).\boxed{\dot U=v_si-Ri^2,\quad U=Li^2/2+q^2/(2C)}.

(b) Interpretation and further calculation

ΔUT=0⇒Ws,T=∫0TRi2dt.\Delta U_T=0\Rightarrow W_{s,T}=\int_0^TRi^2dt.Ws,T=RIp2T/2.\boxed{W_{s,T}=RI_p^2T/2}.

Physical check

The source replaces dissipation each cycle; it does not continually increase stored energy after transients vanish.