$A$ body of mass $1\,kg$ is executing simple harmonic motion. Its displacement $y$ (in $cm$) at $t$ seconds is given by $y = 6\sin(100t + \pi/4)$. Its maximum kinetic energy is ..... $J$.

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

$A$ body of mass $1 \ kg$ is executing simple harmonic motion $(SHM)$. Its displacement $y$ (in $cm$) at time $t$ is given by $y = 6 \sin (100 t + \pi/4) \ cm$. Its maximum kinetic energy is (in $J$)

Position of a $3 \ kg$ mass moving along the $X$-axis is given by $x = 0.3 \cos (\omega t) \ m$. If $K(t)$ denotes the kinetic energy at time $t$,then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is

Find the displacement of a simple harmonic oscillator at which its $PE$ is half of the maximum energy of the oscillator.

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.