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The steady state output of the circuit shown in the figure is given by $y(t)=A(\omega) \sin (\omega t + \phi ( \omega))$. If the amplitude $\mid A (\omega ) \mid =0.25$, then the frequency $\omega$ is

1. $\frac{1}{\sqrt{3} \: R \: C}$
2. $\frac{2}{\sqrt{3} \: R \: C}$
3. $\frac{1}{R \: C}$
4. $\frac{2}{R \: C}$