= Solution
For sinusoidal <steady state> with convention $e^{i\omega t}$, <complex impedance> is the <voltage> <phasor> divided by the current <phasor>. Its modulus is the <amplitude> ratio and its argument is the voltage-current <phase difference>. Since $V_L=L\,dI/dt$,
$$
Z_L=i\omega L,\qquad Z_C=\frac1{i\omega C},\qquad Z_R=R.
$$
The original circuit is a <series RLC circuit> with output across the <inductor>. Its transfer function is
$$
\boxed{H(\omega)=\frac{V_2}{V_1}
=\frac{i\omega L}{R+i(\omega L-1/(\omega C))},\qquad
|V_2|=\frac{|V_1|\omega L}{\sqrt{R^2+(\omega L-1/(\omega C))^2}}.}
$$
For the original <capacitance>, the outputs are \b[$10.15\,\mathrm{mV}$ at $9\,\mathrm{MHz}$ and $4.65\,\mathrm{mV}$ at $10\,\mathrm{MHz}$], a ratio $0.458$. The current <resonance> is $f_0=1/(2\pi\sqrt{LC})=8.967\,\mathrm{MHz}$, so the circuit favors the first transmitter.
The <resistor> <voltage> is in <phase> with the current; the <inductor> <voltage> leads it by $\pi/2$ and the <capacitor> <voltage> lags by $\pi/2$. Large reactive <voltages> therefore nearly cancel in their <phasor> sum. A large <inductor> <voltage> is consistent with a small supply <voltage>: it is \b[<inductor voltage magnification>, not <power gain>] in a passive circuit.
Reducing the <capacitance> raises $f_0$ to $9.452\,\mathrm{MHz}$, between the two transmitters. Direct evaluation gives \b[$7.02\,\mathrm{mV}$ and $7.22\,\mathrm{mV}$], respectively. Their similar detunings give approximately equal <amplitudes>; they are not exactly equal, partly because the numerator of $H$ also depends on <frequency>.
The <two-tone output of a linear resonant circuit>, for coherent monochromatic sources of equal input <amplitude> and arbitrary fixed input <phases>, is
$$
V_{\rm out}(t)=A\cos(2\pi f_A t+\phi_A)+B\cos(2\pi f_B t+\phi_B).
$$
With $A\simeq B=V_o$, this is approximately
$$
\boxed{V_{\rm out}\simeq2V_o
\cos\!\left[\pi(f_B-f_A)t+\frac{\phi_B-\phi_A}{2}\right]
\cos\!\left[2\pi\frac{f_A+f_B}{2}t+\frac{\phi_A+\phi_B}{2}\right].}
$$
Thus a $9.5\,\mathrm{MHz}$ rapid <oscillation> has a <beat> envelope repeating in magnitude at $1\,\mathrm{MHz}$. The signed envelope cosine has <frequency> $0.5\,\mathrm{MHz}$. Unequal <amplitudes> leave a small nonzero envelope minimum; independently drifting transmitter <phases> make the beating drift. No new <frequency> is generated by this ideal linear circuit.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-ia2014-radio-response.png]
{title=Original calculated inductor-voltage gains and illustrative two-tone output, including the circuit's frequency-dependent phase shifts}
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