Question

In $L-C$ oscillation circuit if the frequency of oscillation of charge is $f$, then the frequency of oscillation of magnetic energy is

• A. $f$
• B. $2f$
• C. $\frac{f}{2}$
• D. $4f$

Answer 1

The correct option is B $2f$

We know that charge on the capacitor at any moment is given by

$q=q_0\sin \omega t$ [Here, $\omega =2\pi f$]

$\therefore$ The current in the circuit is

$i=\frac{dq}{dt}=q_0\omega \cos \omega t$

Therefore, energy stored in the inductor at any instant is

$E=\frac{1}{2}L{i}^2$

$=\frac{1}{2}L\times (q_0\omega \cos \omega t{)}^2$

$=\frac{1}{2}L{q}_0^2{\omega }^2\left[{\cos }^2\omega t\right]$

Let, $E_0=\frac{1}{2}L{q}_0^2{\omega }^2$

$\Rightarrow E=E_0{\cos }^2\omega t$


$E=\frac{E_0[1+\cos 2\omega t]}{2}[\because \cos 2\theta =2{\cos }^2\theta -1]$

$=\frac{E_0}{2}(1+\cos 2\omega t)$

$=\frac{E_0}{2}(1+\cos (2\times 2\pi f\times t)$

$=\frac{E_0}{2}(1+\cos (2\pi f^{\prime }\times t)$

$\therefore f^{\prime }=2f$

So, the frequency of oscillation of magnetic energy is $2f$.

Hence, $\left(\text{B}\right)$ is the correct answer.

Alternate solution: We can use spring block analogy for this problem. In this analogy, we compare charge with mass and magnetic energy with kinetic energy block. As we know that if a particle executes simple harmonic motion with a frequency $f$, then $2f$ will be the frequency of kinetic energy. Similarly, frequency of oscillation of magnetic energy will be $2f$.