Question

The following figure shows the displacement versus time graph for two particles $A$ and $B$ executing simple harmonic motions. The ratio of their maximum velocities (respectively) is

• A. $3:1$
• B. $1:3$
• C. $1:9$
• D. $9:1$

Answer 1

The correct option is A $3:1$

For $A$, time period $T_A=16\text{s}$ (distance between two adjacent crests)
For $B$, time period $T_B=2(20-8)=24\text{s}$
(length between the crest and trough shown$=20\text{s}-8\text{s}=12\text{s}$)
Also, amplitudes $a_A=10\text{cm},a_B=5\text{cm}$
Now, $\frac{(V_{\text{max}}{)}_A}{(V_{\text{max}}{)}_B}=\frac{{\omega }_A{a}_A}{{\omega }_B{a}_B}$
$\frac{\left(\frac{2\pi }{T_A}\right)a_A}{\left(\frac{2\pi }{T_B}\right)a_B}=\frac{T_B{a}_A}{T_A{a}_B}=\frac{24\times 10}{16\times 5}=\frac{3}{1}$