Correct Answer is Option (D)

Potential energy is maximum at maximum distance from mean.

Correct Answer is Option (A)

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

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

k = 50$\pi$²

$\approx$ 500

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

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

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

= 5 cm

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

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

= 0.6255

Correct Answer is Option (D)

$E=\frac{1}{2}K{a}^2$

$\frac{3E}{4}=\frac{1}{2}K(a^2-y^2)$

$\Rightarrow$ $\frac{3}{4}\times \frac{1}{2}K{a}^2=\frac{1}{2}K(a^2-y^2)$

$\Rightarrow$ $y^2=a^2-\frac{3a^2}{4}$

$\Rightarrow$ $y=\frac{a}{2}$

Correct Answer is Option (B)

$K=\frac{1}{2}m{\omega }^2(A^2-x^2)$

$=\frac{1}{2}m{\omega }^2\left(A^2-\frac{A^2}{4}\right)$

$=\frac{1}{2}m{\omega }^2\left(\frac{3A^2}{4}\right)$

$K=\frac{3}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)$

Correct Answer is Option (C)

KE = PE

$\frac{1}{2}k(A^2-X^2)=\frac{1}{2}K{X}^2$

$\Rightarrow$ $A^2-X^2=X^2$

$\Rightarrow$ $2X^2=A^2$

$\Rightarrow$ $X^2=\frac{A^2}{\sqrt{2}}$

$\Rightarrow$ $X=\pm \frac{A}{\sqrt{2}}$