QJEEM22Ph11ce: JEE Main 2022 Physics - Class 11 JEEM22Ph11ce Solution: As per given figures, two springs of spring constants and are…

For case (a),

${K_{eq}} = {{2K} \over 3}$

For case (b),

${K_{eq}} = 3K$

$\because$ $T = 2\pi \sqrt {{m \over K}}$

$\therefore$ ${{{T_a}} \over {{T_b}}} = \sqrt {{{{K_b}} \over {{K_a}}}}$

${3 \over {{T_b}}} = \sqrt {{{3K \times 3} \over {2K}}} = {3 \over {\sqrt 2 }}$

${T_b} = \sqrt 2$

$x = 2$