JEE MAIN - Physics (2025 - 29th January Evening Shift - No. 10)
Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1 and k2, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is
$ \sqrt{\frac{k_2}{k_1}} $
$ \sqrt{\frac{k_1}{k_2}} $
$ \frac{k_2}{k_1} $
$ \frac{k_1}{k_2} $
Explanation
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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.
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