Find the angular frequency of the oscillation for the circuit shown in the figure:

CONCEPT:

LC Oscillations:

• We know that a capacitor and an inductor can store electrical and magnetic energy, respectively.
• When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems.
• Let a capacitor and an inductor are connected as shown in the figure.
• Let a capacitor be charged Qo at t = 0 sec.
• The moment the circuit is completed, the charge on the capacitor starts decreasing, giving rise to a current in the circuit.
• The angular frequency of the oscillation is given as,

$\Rightarrow {\omega }_o=\frac{1}{\sqrt{LC}}$

Where L = self-inductance and C = capacitance

• The charge on the capacitor varies sinusoidally with time as,

⇒ Q = Qocos(ωot)

• The current in the circuit at any time t is given as,

⇒ I = Iosin(ωot)

Where Io = maximum current in the circuit

• The relation between the maximum charge and the maximum current is given as,

⇒ Io = ωoQo

CALCULATION:

Given C₁ = C2 = 8 μF = 8 × 10-6 F and L = 0.4 mH = 0.4 × 10-3 H

Where L = self-inductance and C = capacitance

In the given figure the two capacitors are in series so the equivalent capacitance C is given as,

$\Rightarrow \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

$\Rightarrow \frac{1}{C}=\frac{1}{8\times {10}^{-6}}+\frac{1}{8\times {10}^{-6}}$

⇒ C = 4 × 10^-6 C

We know that the angular frequency of the LC oscillation circuit is given as,

$\Rightarrow {\omega }_o=\frac{1}{\sqrt{LC}}$

$\Rightarrow {\omega }_o=\frac{1}{\sqrt{0.4\times {10}^{-3}\times 4\times {10}^{-6}}}$

⇒ ω_o = 25 × 103 rad/sec

• Hence, option 3 is correct.