If the amplitude of a body executing $SHM$ is doubled,what will be its energy?

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

$A$ particle is executing simple harmonic motion with amplitude $A$. The ratio of the kinetic energies of the particle when it is at displacements of $\frac{A}{4}$ and $\frac{A}{2}$ from the mean position is

For a particle executing $S.H.M.$, its potential energy is $8$ times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of $S.H.M.$, the value of '$x$' is

The potential energy of a particle with displacement $X$ is $U(X)$. The motion is simple harmonic,when ($K$ is a positive constant)

$A$ block kept on a frictionless horizontal surface is connected to one end of a horizontal spring of constant $100 \text{ Nm}^{-1}$ whose other end is fixed to a rigid vertical wall. Initially, the block is at its equilibrium position. The block is pulled to a distance of $8 \text{ cm}$ and released. The kinetic energy of the block when it is at a distance of $3 \text{ cm}$ from the mean position is (in $\text{ J}$)