Correct Answer is Option (D)

$K.E=\frac{1}{2}k\left(A^2-d^2\right)$

and $P.E.=\frac{1}{2}k{d}^2$

At mean position $d=0.$ At extremes positions $d=A$

Correct Answer is Option (B)

KEY CONCEPT : The instantaneous kinetic energy of a particle executing $S.H.M.$ is given by

$K=\frac{1}{2}m{a}^2{\omega }^2{\sin }^2\omega t$

$\therefore$ average $K.E.=<K>=<\frac{1}{2}m{\omega }^2{a}^2{\sin }^2\omega t>$

$=\frac{1}{2}m\omega {}^2{a}^2<{\sin }^2\omega t>$

$=\frac{1}{2}m{\omega }^2{a}^2\left(\frac{1}{2}\right)$

$($ as $<{\sin }^2\theta >=\frac{1}{2})$

$=\frac{1}{4}m{\omega }^2{a}^2=\frac{1}{4}m{a}^2{\left(2\pi v\right)}^2$

$($ $\omega =2\pi v)$

or, $<K>={\pi }^{2}m{a}^2{v}^2$

Correct Answer is Option (A)

$K.E.$ of a body undergoing $SHM$ is given by,

$K.E.=\frac{1}{2}m{a}^2{\omega }^2{\cos }^2\omega t,$

$T.E.=\frac{1}{2}m{a}^2{\omega }^2$

Given $K.E.=0.75T.E.$

$\Rightarrow 0.75={\cos }^2\omega t\Rightarrow \omega t=\frac{\pi }{6}$

$\Rightarrow t=\frac{\pi }{6\times \omega }\Rightarrow t=\frac{\pi \times 2}{6\times 2\pi }\Rightarrow t=\frac{1}{6}s$

Correct Answer is Option (A)

At any instant the total energy is

$\frac{1}{2}k{A}^2=$ constant, where $A=$ amplitude

hence total energy is independent of $x.$

Correct Answer is Option (A)

$K.E.=\frac{1}{2}m{\omega }^2\left(a^2-x^2\right)$

When $x=0,$ $K.E$ is maximum and is equal to $\frac{1}{2}m{\omega }^2{a}^2.$