tech99 said:
At resonance the line presents a purely resistive load to the oscillator.
Thanks, yes this is how I finally arrived at the solution. It's given that an open-ended, quarter-wavelength line of length ##2a = 0.25 \ \mathrm{m}## is at resonance. When a capacitance ##C## is then connected across the end of the line, resonance occurs when the line length has been reduced to ##a##. I used the results from the previous analysis to write down\begin{align*}
\dfrac{Z_I}{Z} &= \dfrac{Z_0 \cos{ka} + iZ\sin{ka}}{Z \cos{ka} + iZ_0 \sin{ka}} \\ \\
&= \dfrac{i (Z^2 - Z_0^2) \sin{ka} \cos{ka} + Z_0 Z }{Z^2 \cos^2{ka} + Z_0^2 \sin^2{ka}}
\end{align*}where ##Z_0## is the terminal impedance, ##Z## the (real) line impedance and ##Z_I## the input impedance. The initial quarter-wavelength line condition implies that ##k = 2\pi \ \mathrm{m}^{-1}##, which means ##ka = \pi/4## and ##\sin{ka} \cos{ka} = 1/2##. Given that ##Z_0 = 1/i\omega C##, then ##Z_I## is real if\begin{align*}
i (Z^2 - Z_0^2) \sin{ka} \cos{ka} + Z_0 Z = \dfrac{i}{2} \left(Z^2 + \dfrac{1}{\omega^2 C^2} \right) - \dfrac{iZ}{\omega C}
\end{align*}is real, which is only the case if it is zero. It remains to solve\begin{align*}
Z^2 - \dfrac{2}{\omega C} Z + \dfrac{1}{\omega^2 C^2} = 0
\end{align*}Since ##\omega = ck = 2\pi c##, this becomes
\begin{align*}
Z^2 - \dfrac{1}{\pi c C} Z + \dfrac{1}{4\pi^2 c^2 C^2} = 0
\end{align*}Inserting the value for ##C## gives, to my surprise, the correct answer.
