Answer
(a) $535.6Hz$
(b) decreased
Work Step by Step
(a) We know that
$Z=\sqrt{R^2+(X_L-X_C)^2}$
$\implies 2000=\sqrt{(1000)^2+(\omega L-\frac{1}{\omega C})^2}$
This simplifies to:
$\frac{\omega^2LC-1}{\omega C}=732$
$\implies \omega^2 515\times 10^{-3}\times 252\times 10^{-6}-1=1732\omega\times 252\times 10^{-6}$
$\implies \omega=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
This simplifies to:
$\omega=3365.8s^{-1}$ or $-1.34s^{-1}$, but angular frequency is a positive quantity.
Now $f=\frac{\omega}{2\pi}$
We plug in the known values to obtain:
$f=\frac{3365.8}{2\pi}$
$f=535.6Hz$
(b) We know that the resonance frequency is given as
$f_{\circ}=\frac{1}{2\pi \sqrt{LC}}$
We plug in the known values to obtain:
$f_{\circ}=\frac{1}{2\pi \sqrt{129780\times 10^{-9}}}$
$\implies f_{\circ}=14Hz$
Thus, to reduce the impedance of the circuit, the frequency is required to be changed so that it reaches to its resonance frequency. At resonance frequency $f_{\circ}=14Hz$ and hence the frequency is decreased.
