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 $s$)

Show that the motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is simple harmonic with a period of $\frac{2\pi}{\omega}$.

$A$ particle executing simple harmonic motion along the $Y$-axis has its motion described by the equation $y = A \sin(\omega t) + B$. The amplitude of the simple harmonic motion is

$A$ body is executing simple harmonic motion of amplitude $a$ and period $T$ about the equilibrium position $x=0$. Large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval $x$ to $x+|dx|$ is highest at

When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=kx^2$,it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$,as can be seen easily using dimensional analysis. However,the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $kx^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $m$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4$ $(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x| \geq X_0$ (see figure).
$1.$ If the total energy of the particle is $E$,it will perform periodic motion only if
$(A)$ $E < 0$
$(B)$ $E > 0$
$(C)$ $V_0 > E > 0$
$(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $A$,the time period $T$ of this particle is proportional to
$(A)$ $A \sqrt{\frac{m}{\alpha}}$
$(B)$ $\frac{1}{A} \sqrt{\frac{m}{\alpha}}$
$(C)$ $A \sqrt{\frac{\alpha}{m}}$
$(D)$ $A \sqrt{\frac{\alpha}{m}}$
$3.$ The acceleration of this particle for $|x|>X_0$ is
$(A)$ proportional to $V_0$
$(B)$ proportional to $\frac{V_0}{mX_0}$
$(C)$ proportional to $\sqrt{\frac{V_0}{mX_0}}$
$(D)$ zero
Give the answer for questions $1, 2$ and $3$.

$A$ particle is executing simple harmonic motion with an instantaneous displacement $x = A \sin^2(\omega t - \frac{\pi}{4})$. The time period of oscillation of the particle is