From the potential energy curve,

$${U_{\max }} = {1 \over 2}k{A^2}$$

$$10 = {1 \over 2}k{(2)^2}$$

$$\Rightarrow$$ k = 5 N/m

The length of the simple pendulum, L = 4 m

Time period of spring,

$$T = 2\pi \sqrt {{k \over m}} $$

Time period of simple pendulum,

$$T = 2\pi \sqrt {{l \over g}} $$

The time period of simple pendulum is same as the time period of the spring oscillation.

$$\Rightarrow$$ $$2\pi \sqrt {{l \over g}} $$ = $$2\pi \sqrt {{k \over m}} $$

Substituting the values in the above equations, we get

$$2\pi \sqrt {{4 \over g}} $$ = $$2\pi \sqrt {{5 \over 5}} $$

g = 4m/s²

$$\therefore$$ The acceleration due to the gravity on the planet is 4 m/s².

In S.H.M. total mechanical energy remains constant and also = = $${{1 \over 4}}$$KA² (for 1 time period)

Potential energy is maximum at maximum distance from mean.

$$T = 2\pi \sqrt {{m \over k}} $$

$$0.2 = 2\pi \sqrt {{{0.5} \over k}} $$

k = 50$$\pi$$²

$$ \approx $$ 500

x = A sin ($$\omega$$t + $$\phi$$)

= 5 cm sin $$\left( {{{\omega T} \over 4} + 0} \right)$$

= 5 cm sin $$\left( {{\pi \over 2}} \right)$$

= 5 cm

$$PE = {1 \over 2}k{x^2}$$

$$ = {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}$$

= 0.6255