The maximum potential energy of a block executing $S.H.M.$ is $E$ and amplitude of oscillation is $A$. The kinetic energy of the block at amplitude $A/2$ is

$A$ particle of mass $m$ is executing simple harmonic motion about its mean position. If $A$ is the amplitude and $T$ is the period of $S$.$H$.$M$.,then the total energy of the particle is

The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = k x^{2} / 2$,where $k$ is the force constant of the oscillator. For $k = 0.5 \; N m^{-1}$,the graph of $V(x)$ versus $x$ is shown in the figure. Show that a particle of total energy $1 \; J$ moving under this potential must 'turn back' when it reaches $x = \pm 2 \; m$.

$A$ particle is executing $S.H.M.$ of amplitude $A$. When the potential energy of the particle is half of its maximum value during the oscillation,its displacement from the equilibrium position is

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

If the frequency of a simple harmonic motion is $f$,what is the frequency of its kinetic energy?