$A$ body is performing $S.H.M.$ of amplitude $A$. The displacement of the body from a point where kinetic energy is maximum to a point where potential energy is maximum,is

In a simple harmonic motion,when the displacement is one-half the amplitude,what fraction of the total energy is kinetic?

$A$ particle of mass $m$ is executing $S.H.M.$ about the origin on the $x$-axis with frequency $f = \frac{\sqrt{Ka}}{\pi m}$, where $K$ is a constant and $a$ is the amplitude of $S.H.M.$ If $x$ is the displacement of the particle at time $t$, the potential energy of the particle will be:

$A$ body of mass $m$ is executing $SHM$ with amplitude $a$. When its displacement $x = 1$ unit,the force is $b$. What will be its maximum kinetic energy?

In $S.H.M.$,the displacement of a particle at an instant is $Y = A \cos 30^{\circ}$,where $A = 40 \ cm$ and kinetic energy is $200 \ J$. If the force constant is $1 \times 10^{x} \ N/m$,then $x$ will be $(\cos 30^{\circ} = \sqrt{3}/2)$.

Starting from the origin,a body oscillates simple harmonically with a period of $2 \ s$. After what time will its kinetic energy be $75\%$ of the total energy?