$A$ particle free to move along the $x$-axis has potential energy given by $U(x) = k[1 - \exp(-x^2)]$ for $-\infty \le x \le +\infty$,where $k$ is a positive constant of appropriate dimensions. Then:

The radius of the circle,the period of revolution,the initial position,and the sense of revolution are indicated in the figure. The $y$-projection of the radius vector of the rotating particle $P$ is:

The motion of a particle executing simple harmonic motion is described by the displacement function,$x(t) = A \cos (\omega t + \phi)$. If the initial $(t = 0)$ position of the particle is $1 \; cm$ and its initial velocity is $\omega \; cm/s$,what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \; s^{-1}$. If instead of the cosine function,we choose the sine function to describe the $SHM$: $x = B \sin (\omega t + \alpha)$,what are the amplitude and initial phase of the particle with the above initial conditions?

Equations $y_1 = A \sin \omega t$ and $y_2 = \frac{A}{2} \sin \omega t + \frac{A}{2} \cos \omega t$ represent $S.H.M.$ The ratio of the amplitudes of the two motions is

$A$ particle executing simple harmonic motion of amplitude $5\, cm$ has a maximum speed of $31.4\, cm/s$. The frequency of its oscillation is ..... $Hz$.

The equation of $SHM$ of a particle is given as $2 \frac{d^2x}{dt^2} + 32x = 0$,where $x$ is the displacement from the mean position of rest. The period of its oscillation (in seconds) is