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(C) The angular frequencies of the two oscillators are:$\omega_1 = \sqrt{\frac{k}{m_1}} = \sqrt{\frac{1200}{3.0}} = 20 \ rad/s$$\omega_2 = \sqrt{\frac

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$\frac{3\pi}{40} \ s$

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(C) The angular frequencies of the two oscillators are:$\omega_1 = \sqrt{\frac{k}{m_1}} = \sqrt{\frac{1200}{3.0}} = 20 \ rad/s$$\omega_2 = \sqrt{\frac{k}{m_2}} = \sqrt{\frac{1200}{27}} = \sqrt{\frac{400}{9}} = \frac{20}{3} \ rad/s$Both particles are released from the same initial displacement $A = 10 \ cm$. Their positions as a function of time are $x_1(t) = A \cos(\omega_1 t)$ and $x_2(t) = A \cos(\omega_2 t)$.They are side by side when $x_1(t) = x_2(t)$,which implies $\cos(20t) = \cos(\frac{20}{3}t)$.This occurs when $20t = 2n\pi \pm \frac{20}{3}t$. For the first time after $t=0$,we take the negative sign:$20t = 2\pi - \frac{20}{3}t$$(20 + \frac{20}{3})t = 2\pi$$(\frac{60+20}{3})t = 2\pi$$\frac{80}{3}t = 2\pi$$t = \frac{6\pi}{80} = \frac{3\pi}{40} \ s$.

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