If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4v^2 = 25 - x^2$, then the time period is

The function $\sin^2(\omega t)$ represents:

The figure shows the circular motion of a particle. The radius of the circle is $B$,the period is $30 \ s$,the sense of revolution is clockwise,and the initial position at $t = 0$ is at an angle of $45^\circ$ (or $\pi/4$ radians) with the positive $y$-axis in the first quadrant. The simple harmonic motion of the $x$-projection of the radius vector of the rotating particle $P$ is:

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 . . . . . . .

The displacement of a particle is represented by the equation $y = \sin^3 \omega t$. The motion is

$A$ particle performs $SHM$ on the $x-$axis with a time period of $0.5 \, s,$ such that its velocity is zero at $x = -3 \, cm$ and at $x = 9 \, cm$. It was located at $x = 0$ and moving in the negative $x-$direction at $t = 0$. The equation of $SHM$ of the particle is: