Two particles A and B execute SHMs of periods | vedclass

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(C) The time taken by particle $A$ to complete two oscillations is $t = 2T$.For particle $A$,the angular frequency is $\omega_A = \frac{2\pi}{T}$.The

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$\frac{4\pi}{3}$

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(C) The time taken by particle $A$ to complete two oscillations is $t = 2T$.For particle $A$,the angular frequency is $\omega_A = \frac{2\pi}{T}$.The phase of particle $A$ at time $t = 2T$ is $\phi_A = \omega_A t = \left(\frac{2\pi}{T}\right)(2T) = 4\pi$.For particle $B$,the angular frequency is $\omega_B = \frac{2\pi}{(3T/2)} = \frac{4\pi}{3T}$.The phase of particle $B$ at time $t = 2T$ is $\phi_B = \omega_B t = \left(\frac{4\pi}{3T}\right)(2T) = \frac{8\pi}{3}$.The phase difference between them is $\Delta\phi = |\phi_A - \phi_B| = |4\pi - \frac{8\pi}{3}| = |\frac{12\pi - 8\pi}{3}| = \frac{4\pi}{3}$.

Two particles $A$ and $B$ execute $SHMs$ of periods $T$ and $\frac{3T}{2}$. If they start from the mean position,then the phase difference between them,when the particle $A$ completes two oscillations,will be

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