To determine the conditions for sustained oscillations in the given circuit, we analyze the feedback network and apply the Barkhausen Criterion.

1. Identify the Feedback Network

The circuit consists of a non-inverting amplifier and an RC feedback network connected to the non-inverting terminal (\(+\)). The feedback network is a two-stage Lead-Lag network (common in Wien Bridge oscillators), consisting of:

• A series combination of capacitor \(C\) and a shunt resistor \(R\).
• Followed by a series resistor \(R\) and a shunt capacitor \(C\).

2. Calculate the Feedback Factor (\(\beta\))

The feedback factor \(\beta(s)\) is the transfer function from the output of the op-amp (\(V_{out}\)) to the non-inverting input (\(V_+\)). Using circuit analysis:

$$\beta(s) = \frac{V_+}{V_{out}} = \frac{1}{sRC + 3 + \frac{1}{sRC}}$$

Substituting \(s = j\omega\):

$$\beta(j\omega) = \frac{1}{3 + j\left(\omega RC - \frac{1}{\omega RC}\right)}$$

3. Apply the Barkhausen Criterion

For sustained oscillations, the loop gain must satisfy \(A\beta = 1\), where \(A\) is the gain of the non-inverting amplifier. This requires two conditions:

• Phase Condition: The phase shift of the loop must be \(0^\circ\). This occurs when the imaginary part of the denominator of \(\beta(j\omega)\) is zero: $$\omega RC - \frac{1}{\omega RC} = 0 \implies \omega^2 = \frac{1}{(RC)^2} \implies \omega = \frac{1}{RC}$$ The frequency of oscillation is: $$f = \frac{\omega}{2\pi} = \frac{1}{2\pi RC}$$
• Magnitude Condition: At this frequency, \(\beta = \frac{1}{3}\). Thus, we need a gain \(A = 3\). For a non-inverting amplifier: $$A = 1 + \frac{R_f}{R_{in}} = 1 + \frac{R_1}{R}$$ Setting the gain to 3: $$1 + \frac{R_1}{R} = 3 \implies \frac{R_1}{R} = 2 \implies R_1 = 2R$$

Conclusion

Based on the analysis, for sustained oscillations, the respective values are:

• Resistor \(R_1\): \(2R\)
• Frequency of oscillation: \(1/(2\pi RC)\)

The correct option is \(2R\) and \(1/(2\pi RC)\).