
${U}_{\max }=\frac{1}{2} {kA}^{2} \Rightarrow 10=\frac{1}{2} {k}(2)^{2}$
$\Rightarrow {k}=5$
Now ${T}_{\text {spring }}={T}_{\text {pendulum }}$
$2 \pi \sqrt{\frac{5}{5}}=2 \pi \sqrt{\frac{4}{g}}$
$\Rightarrow 1=\sqrt{\frac{4}{g}} \Rightarrow g=4 \text { on planet }$
