Two bodies A and B of equal mass are suspended from two massless springs of spring constant \({k}_{1}\) and \({k}_{2}\),…

the maximum velocity for a simple harmonic oscillator is given by:

\(V=A\omega\)

The angular frequency \(\omega\) of a mass-spring system is related to the spring constant k and the mass m by:

\(\omega =\sqrt{\frac{k}{m}}\)

Since the bodies have equal amplitudes, therefore,

\(\frac{{V}_{1}}{{V}_{2}}=\frac{{\omega }_{1}}{{\omega }_{2}}\\ \frac{{V}_{1}}{{V}_{2}}=\frac{\sqrt{\frac{{k}_{1}}{m}}}{\sqrt{\frac{{k}_{2}}{m}}}\\ \frac{{V}_{1}}{{V}_{2}}=\sqrt{\frac{{k}_{1}}{{k}_{2}}}\)
Option (c) is correct.