If the displacement of a particle executing simple harmonic motion is given by $x = 0.5 \cos (125.6 t)$,then the time period of oscillation of the particle is nearly. (Here $x$ is displacement in metre and $t$ is time in second) (in $s$)

$A$ particle of mass $0.1 \ kg$ is executing simple harmonic motion of amplitude $0.1 \ m$. When the particle passes through the mean position,its kinetic energy is $8 \times 10^{-3} \ J$. If the initial phase is $45^{\circ}$,the equation of its motion is (Assume $x(t)$ as the position of the particle at time $t$)

The equation of motion of a particle is given by $x = a \sin(50t + \frac{\pi}{3}) \text{ cm}$. The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$. The $t_1$ and $t_2$ respectively are . . . . . . .

Write the force law for $SHM$ and obtain the formula for the period of an $SHM$ particle.

The motion of a particle varies with time according to the relation $y = a(\sin \omega t + \cos \omega t)$.

The figure shows the circular motion of a particle. The radius of the circle,the period,the sense of revolution,and the initial position are indicated in the figure. The simple harmonic motion of the $x$-projection of the radius vector of the rotating particle $P$ is