QJEEM25Ph702b: JEE Main 2025 Physics - Class 11 JEEM25Ph702b Solution: Two bodies A and B of equal mass are suspended from two…

We know, for an oscillation,

${V_{\max }} = wA$ .... (1)

where, A = amplitude

w = frequency of oscillation

and $w = \sqrt {{k \over m}}$ .... (2)

from (1) and (2), ${V_{\max }} = \sqrt {{k \over m}} A$

$\Rightarrow {V_{\max }} \propto \sqrt k$ (As m and A are same (or constant))

$\Rightarrow {{{{\left( {{V_{\max }}} \right)}_A}} \over {{{\left( {{V_{\max }}} \right)}_B}}} = \sqrt {{{{k_1}} \over {{k_2}}}}$

Hence, option 2 is correct.