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(B) Let the amplitude be $A$. The angular frequencies are $\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{3}$ and $\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi

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$2: 1$

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(B) Let the amplitude be $A$. The angular frequencies are $\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{3}$ and $\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{6} = \frac{\pi}{3}$.Since they start from the origin,their displacements are $x_1 = A \sin(\omega_1 t)$ and $x_2 = A \sin(\omega_2 t)$.When they meet,$x_1 = x_2$,so $\sin(\omega_1 t) = \sin(\omega_2 t)$.This implies $\omega_1 t = \pi - \omega_2 t$ (for the first meeting point),so $t = \frac{\pi}{\omega_1 + \omega_2} = \frac{\pi}{\frac{2\pi}{3} + \frac{\pi}{3}} = 1 \ s$.The velocity of a particle in $SHM$ is $v = A\omega \cos(\omega t)$.At $t = 1 \ s$,$v_P = A(\frac{2\pi}{3}) \cos(\frac{2\pi}{3} \cdot 1) = A(\frac{2\pi}{3})(-\frac{1}{2}) = -\frac{A\pi}{3}$.At $t = 1 \ s$,$v_Q = A(\frac{\pi}{3}) \cos(\frac{\pi}{3} \cdot 1) = A(\frac{\pi}{3})(\frac{1}{2}) = \frac{A\pi}{6}$.The ratio of the magnitudes of velocities is $|v_P| : |v_Q| = \frac{A\pi}{3} : \frac{A\pi}{6} = 2 : 1$.

Two particles $P$ and $Q$ start from the origin and execute simple harmonic motion along the $X$-axis with the same amplitude but with periods $3 \ s$ and $6 \ s$,respectively. The ratio of the velocities of $P$ and $Q$ when they meet is

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