$A$ ball is released from height $h$ which makes a perfectly elastic collision with the ground. The frequency of the periodic vibratory motion is ($g$ = acceleration due to gravity).

Give the relation between periodic time and angular frequency.

The displacement of a particle executing linear $S.H.M.$ is given by $x = 0.25 \sin(11t + 0.5) \ m$. The period of $S.H.M.$ is (take $\pi = \frac{22}{7}$):

Fill in the blanks:
$1.$ In $SHM$,......... quantities are always positive.
$2.$ $A$ periodic motion that obeys the force law ......... is only a simple harmonic motion.
$3.$ $A$ $SHO$ with a periodic time of $2 \ s$ starts its oscillation from the lower end of its path of motion; its phase will be .......... at time $t = 2 \ s$.

The displacement of a particle executing $S.H.M.$ is given by $x = 0.01 \sin 100 \pi(t + 0.05)$. The time period is ........ $s$.

The figures correspond to two circular motions. The radius of the circle,the period of revolution,the initial position,and the sense of revolution (i.e.,clockwise or anti-clockwise) are indicated on each figure. Obtain the corresponding simple harmonic motions of the $x$-projection of the radius vector of the revolving particle $P$,in each case.