The following figure shows the displacement versus time graph for two particle s A and B executing simple harmonic motions. The ratio of their maximum velocities is

For A, time period T_A = 16 s (distance between two adjacent crests)

For B, time period $T_B=2(20-8)=24\mathrm{s}$ (length between the crest and trough shown $=20\mathrm{s}-8\mathrm{s}=12\mathrm{s}\text{ ) }$

Also, amplitudes $a_A=10\mathrm{cm},a_B=5\mathrm{cm}$

$\text{ Now, }\frac{{\left(V_{max}\right)}_A}{{\left(V_{min}\right)}_B}=\frac{{\omega }_A{a}_A}{{\omega }_B{a}_B}=\frac{\left(\frac{2\pi }{T_A}\right)a_A}{\left(\frac{2\pi }{T_B}\right)a_B}=\frac{T_B{a}_A}{T_A{a}_B}=\frac{24\times 10}{16\times 5}=\frac{3}{1}$