Two particles P and Q start from the origin and execute simple harmonic motion along X-axis with the same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is
(A) 1:2
(B) 2:1
(C) 2:3
(D) 3:2

Hint: In this question, we need to determine the ratio of the velocity of the particles P and Q such that the time period of the particles executing simple harmonic motion is 3 seconds and 6 seconds respectively. For this, we will follow the relation between the velocity of the particle at the equilibrium point, frequency and the amplitude of the particles executing simple harmonic motion.

Complete step by step answer:
According to the question, the time period of the particles P and Q executing simple harmonic motion is 3 seconds and 6 seconds respectively. So, we can say that when the particle P completes one cycle with respect to the equilibrium point then, the particle Q completes only half of its cycle. So, they will meet at the origin only.
In the simple harmonic motion, the velocity of the particle executing simple harmonic motion is maximum at the equilibrium point. The maximum velocity of the particle in a simple harmonic motion is given as the product of the amplitude of the particle and the frequency. Mathematically, $v = \omega A$ where, ‘v’ is the linear velocity of the particle, $\omega $ is the frequency and ‘A’ is the amplitude of the particle in the simple harmonic motion.
Let the velocity of the particle P be ${v_P}$ , and the velocity of the particle Q be ${v_Q}$.
Then, the velocity of the particle P at the origin is given as
$
{v_P} = \omega A \\
\Rightarrow{v_P} = \dfrac{{2\pi }}{{{T_P}}} \times A - - - - (i) \\
 $
Similarly, the velocity of the particle Q at the origin is given as
$
{v_Q} = \omega A \\
\Rightarrow{v_Q} = \dfrac{{2\pi }}{{{T_Q}}} \times A - - - - (ii) \\
$
Dividing the equations (i) and (ii) to evaluate the ratio of the velocity of the particles when they meet (at origin).
$
\dfrac{{{v_P}}}{{{v_Q}}} = \left( {\dfrac{{\dfrac{{2\pi }}{{{T_P}}} \times A}}{{\dfrac{{2\pi }}{{{T_Q}}} \times A}}} \right) \\
\Rightarrow\dfrac{{{v_P}}}{{{v_Q}}} = \dfrac{{{T_Q}}}{{{T_P}}} \\
$
Substituting the values of the time period in the above equation, we get
$
\dfrac{{{v_P}}}{{{v_Q}}} = \dfrac{{{T_Q}}}{{{T_P}}} \\
\Rightarrow\dfrac{{{v_P}}}{{{v_Q}}} = \dfrac{6}{3} \\
\therefore\dfrac{{{v_P}}}{{{v_Q}}} = \dfrac{2}{1} \\
$
Hence, the ratio of the velocity of the particles P and Q is 2:1.Thus,option B is correct.

Note: Here, in the question, it is given that the amplitude of the particles are the same, so we have taken it so. Moreover, it should be made sure that where the particles meet on the x-axis to proceed with the solution.