Mark the wrong statement.

$A$ particle is performing $SHM$ according to the equation $x = (3\, cm) \sin \left( \frac{2\pi t}{18} + \frac{\pi}{6} \right)$ where $t$ is in seconds. The distance travelled by the particle in $36\, s$ is ..... $cm$.

$A$ particle executes simple harmonic motion with an amplitude '$A$'. The distance travelled by it in one periodic time is

Who decides the characteristics of $SHM$?

$A$ particle is executing simple harmonic motion with an amplitude of $2 \,m$. The difference in the magnitudes of its maximum acceleration and maximum velocity is $4$. The time-period of its oscillation and its velocity when it is $1 \,m$ away from the mean position are respectively:

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 $3$.