Question

Two particles P and Q start from origin and execute S.H.M. along the 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 are at mean position is
A. 1:2
B. 2:1
C. 2:3
D. 3:2

Answer 1

Hint: Velocity at the mean position of a S.H.M. is product of the amplitude and the angular frequency. Frequency is inverse of the time period of oscillation. Use these to calculate velocities for P and Q and divide them to find the ratio.

Complete answer:
We have been given that the amplitudes of the two S.H.M. ( simple harmonic motion), P and Q, are equal and let that be $A$. Let the time periods and angular frequencies of the two motions be $({t_1},{t_2})$ and $({\omega _1},{\omega _2})$ respectively where, ${t_1} = 3s;{t_2} = 6s$. Angular frequency and time period are related as follows:
$\omega = \dfrac{{2\pi }}{t}.$ Therefore, ${\omega _1} = \dfrac{{2\pi }}{{{t_1}}} = \dfrac{{2\pi }}{3};{\omega _2} = \dfrac{{2\pi }}{{{t_2}}} = \dfrac{{2\pi }}{6}$.
The velocity at the mean position of the motion is given as the product of amplitude and angular frequency of the motion.
For P, velocity, ${v_1} = A{\omega _1} = \dfrac{{2\pi A}}{3}$ and for Q, ${v_2} = A{\omega _2} = \dfrac{{2\pi A}}{6}$.
Therefore,$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{\dfrac{2\pi A}{3}}{\dfrac{2\pi A}{6}} = \dfrac{2}{1} = 2:1$. So option B is correct.

Note: Option A is the exact reciprocal of the correct answer. So be careful while reading the question and substitute the time periods accordingly. If you consider the ratio of Q versus P then you will get a ratio of 1:2 and you will pick option A as the correct answer and lose the problem.