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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

Vedclass · Accepted Answer

$2: 1$

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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 both start from the origin at $t=0$, 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 $\sin(\omega_1 t) = \sin(\omega_2 t)$.For the first meeting after $t=0$, $\omega_1 t = \pi - \omega_2 t$, which gives $t = \frac{\pi}{\omega_1 + \omega_2} = \frac{\pi}{\frac{2\pi}{3} + \frac{\pi}{3}} = 1 \ s$.The velocity of a particle in $SHM$ is $v = A\omega \cos(\omega t)$.The ratio of velocities is $\frac{v_P}{v_Q} = \frac{A\omega_1 \cos(\omega_1 t)}{A\omega_2 \cos(\omega_2 t)} = \frac{(2\pi/3) \cos(2\pi/3 \cdot 1)}{(\pi/3) \cos(\pi/3 \cdot 1)} = \frac{2 \cos(120^\circ)}{\cos(60^\circ)} = \frac{2(-1/2)}{1/2} = -2$.The magnitude ratio is $2:1$.

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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