The Hartley oscillator is an LC (inductor-capacitor) oscillator used in radio frequency applications. The oscillation frequency is given by the formula:

\(f = \frac{1}{2\pi\sqrt{L_{\text{eq}}C}}\)

where \(f\) is the frequency of oscillation, \(L_{\text{eq}}\) is the equivalent inductance, and \(C\) is the capacitance.

In this problem, it is given that the equivalent inductance \(L_{\text{eq}}\) is doubled. Let's see the effect on the frequency:

1. Initial frequency, \(f_1 = \frac{1}{2\pi\sqrt{L_{\text{eq}}C}}\)
2. After doubling the inductance, the new equivalent inductance becomes \(2L_{\text{eq}}\).
3. The new frequency, \(f_2 = \frac{1}{2\pi\sqrt{2L_{\text{eq}}C}}\)

Let's calculate \(f_2\) in terms of \(f_1\):

\(f_2 = \frac{1}{2\pi\sqrt{2L_{\text{eq}}C}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2\pi\sqrt{L_{\text{eq}}C}} = \frac{1}{\sqrt{2}} f_1\)

This shows that the frequency \(f_2\) is reduced by a factor of \(\sqrt{2}\) compared to \(f_1\).

Hence, the frequency is reduced by \(\sqrt{2}\) when the equivalent inductance is doubled. Therefore, the correct answer is:

Reduced by √2