The charge in an LC circuit with negligible r | vedclass

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(B) The given differential equation is $\frac{d^2q}{dt^2} + \omega^2q = 0$,where $\omega^2 = 16\pi^2$. Thus,$\omega = 4\pi \, rad/s$. The general solu

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$12$

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(B) The given differential equation is $\frac{d^2q}{dt^2} + \omega^2q = 0$,where $\omega^2 = 16\pi^2$. Thus,$\omega = 4\pi \, rad/s$. The general solution for the charge $q(t)$ is $q(t) = q_0 \cos(\omega t + \phi)$. Given that at $t = 0$,the charge is maximum $(q_0 = 24\,\mu C)$,we have $\phi = 0$. So,$q(t) = 24 \cos(4\pi t)$. At $t = \frac{1}{12}\,s$,the charge is $q = 24 \cos(4\pi 	imes \frac{1}{12}) = 24 \cos(\frac{\pi}{3})$. Since $\cos(\frac{\pi}{3}) = 0.5$,we get $q = 24 	imes 0.5 = 12\,\mu C$.

The charge in an $LC$ circuit with negligible resistance oscillates as given by the equation $\frac{d^2q}{dt^2} + 16\pi^2q = 0$. If the charge is maximum equal to $24\,\mu C$ at $t = 0$,find the charge at $t = \frac{1}{12}\,s$ in $\mu C$.

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