Oscillations and Waves: Winter-2026
HW 3a (SOLUTION): Due W3 D3

1. Q factor of a Resonance Circuit S1 5479S

   We define a “quality factor”, \(Q\equiv\frac{\omega_0}{2\beta}\), and use it as a measure of cycles in a free (undriven), damped oscillator before the oscillation decays to some smaller amplitude. The larger the number of cycles, the larger the \(Q\). Now let's see what this quantity translates to in a driven, damped oscillator. Use the example of charge amplitude\(|q|\) for a series LRC circuit.

   1. Show that at both frequencies \(\omega=\omega_0 +\beta\) and \(\omega=\omega_0-\beta\), the magnitude of the charge response \(|q|\) is \(\frac{|q|_{max}}{\sqrt{2}}\).
      (Hint -- this is a 3-line calculation. You don't need the definition of \(|q|\) in terms of \(L,\,R,\,C\)).

      The definition of \(\omega_+\equiv\omega_0+\beta\) is \(\frac{E(\omega_+)}{E_{max}}=\frac{1}{2}\)
      Because \(E=\frac{|q|^2}{2C}\), it follows that \(\frac{|q|^2(\omega_+)/2C}{|q|^2_{max}/2C}=\frac{1}{2}\)
      Cancel terms and take the square root, \(\frac{|q|(\omega_+)}{|q|_{max}}=\frac{1}{\sqrt{2}}\)]] Same argument for \(\omega_-\equiv\omega_0-\beta\)

   2. At these frequencies, how large is the energy in the capacitor compared to the maximal energy at \(\omega\approx\omega_0\)?

      The energy in a capacitor is proportional the the charge squared, so at those frequencies, the energy is half its value compared to at resonance.

   3. Given the above, operationally, how would you measure the \(Q\) of a resonant circuit? What is \(Q\) for your circuit?

      You didn't measure \(|q|\), you measured \(|dq/dt|\), but it is proportional to \(|q|\), so find the frequencies where the value of \(|dq/dt|\) falls to \(\frac{1}{\sqrt2}\approx0.7\) of its maximum value. (At these frequenciees, the value of \(|q|^2\) drop to \(\frac{1}{2}\) its maximum value, and hence defines \(Q\)). Find the difference between those two values. That difference corresponds to the FWHM of the energy curve. That is, if you'd plotted \(|q|^2\) and found the frequencies where \(|q|^2\) drop to \(\frac{1}{2}\) its maximum value, you'd get the same result.
      

      For your circuit, \(Q\approx20\)

   The graphs below may help with a visual feel for the quantities discussed above:

   

2. Steady State Solutions for the LRC Circuit S1 5479S

   1. Write the equation of motion governing the charge on the capacitor in a series LRC circuit driven by an external sinusoidal voltage. Identify all parameters in your equation.

      The equation is found from adding thet voltages across the inductor, capacitor, and resistor and setting them equal to the driving voltage: \(V_C+V_L+V_R=V_{ext}\), where \(V_C=\frac{q}{C}\,(q=\) charge and \(C=\) capacitance), \(V_R=IR \,(I=\frac{dq}{dt}=\) current, \(R=\) resistance), and \(V_L=L\frac{dI}{dt}\,(L=\) inductance), \(V_0\) is the amplitude of the external driving voltage. Define \(\frac{1}{LC}\equiv\omega_0^2\); \(\frac{R}{L}\equiv2\beta\) and obtain:

      \[\ddot{q}+2\beta \dot{q}+\omega_0^2 q=\frac{V_0 e^{i\omega t}}{L}\]

   2. Find a steady-state solution (or particular solution) for the current in the circuit. After what time is the steady state the only relevant part of the solution, i.e., after what time has the transient solution decayed for the circuit you are working with?

      The steady state solution is the particular solution. It becomes the only part of the solution after the time longer than \(1/\beta\), when the transient has decayed away. Make the ansatz that \(q(t)=|q|e^{i\phi_q}e^{i\omega t}\) where \(|q|\) is a real number that is the amplitud of the charrge oscillations and \(\phi_q\) is the phase relative to the driving voltage, and substitute: \[-\omega^2|q|e^{i(\omega t +\phi_q)}+2\beta i \omega|q|e^{i(\omega t+\phi_q)}+\omega_0^2|q|e^{i(\omega t+\phi_q)}=\frac{V_0e^{i\omega t}}{L}\] Cancel the time dependence, which proves the ansatz works. Solve for the unknowns: \[|q|e^{i\phi_q}=\frac{V_0/L}{(-\omega^2+2\beta i\omega+\omega_0^2)} \] Rewrite the right-hand side in polar form (class exercise): \[|q|e^{i\phi_q}=\frac{V_0/L}{\sqrt{{(\omega_0^2-\omega^2)+(2\beta\omega)^2}}}e^{-i\theta} \] where \(\tan{\theta}=\frac{2\beta\omega}{\omega_0^2-\omega^2}\). Identify the magnitudes and phases: \[|q|=\frac{V_0/L}{\sqrt{(\omega_0^2-\omega^2)+(2\beta\omega)^2}}\] \[\tan{\phi_q}=\frac{-2\beta\omega}{\omega_0^2-\omega^2}\] Differentiate to find current: \[I(t)=\dot{q}(t)=i\omega q(t)=\omega e^{i\pi/2}q(t)\] so we can write current in the form \[I(t)=Re[|I|e^{i(\omega t+\phi_I)}]\] with \[|I|=\omega|q|=\frac{\omega V_0/L}{\sqrt{(\omega_0^2-\omega^2)^2+(2\beta\omega)^2}}\] \[\phi_I=\frac{\pi}{2}+\phi_q\]

   3. Find the steady state solution for \(\frac{dI}{dt}\) in this circuit.

      Differentiate to find \(\frac{dI}{dt}\): \(\dot{I}(t)=\ddot{q}(t)=-\omega^2q(t)=\omega^2e^{i\pi}q(t)\)

      So we can rewrite the derivative in this form: \(\dot{I}(t)=Re[\,|\dot{I}|\,e^{i(\omega t+\phi_{\dot{I}})}]\) with \[|\dot{I}|=\frac{\omega^2 V_0/L}{\sqrt{{(\omega_0^2-\omega^2)+(2\beta\omega)^2}}}\] \[\phi_{\dot{I}}=\pi+\phi_q\]