The springs in figure. $A$ and $B$ are identical but length in $A$ is three times that in $B$. The ratio of period $T_A/T_B$ is
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$\mathrm{K}_{1} \ell_{1}=\mathrm{K}_{2} \ell_{2}$
let spring constant of spring in figure $A$ is $K.$ now spring constant of each spring in figure $B$ is $3 \mathrm{K}$
in figure, all springs are in parallel so $(K_{eq } ) _ { B }$ $=9 \mathrm{K}$
time period $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$
$\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{K}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{A}}}}=3$
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