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 $(\cos 30^{\circ} = \sqrt{3}/2)$.

For a particle performing $S.H.M.$,the total energy is '$n$' times the kinetic energy,when the displacement of a particle from the mean position is $\frac{\sqrt{3}}{2} A$,where $A$ is the amplitude of $S.H.M.$ The value of '$n$' is

$A$ particle is executing Simple Harmonic Motion $(SHM)$. The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ 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:

$A$ particle of mass $m$ performs $SHM$ along a straight line with frequency $f$ and amplitude $A.$