The displacement $x$ (in metre) of a particle in simple harmonic motion is related to time $t$ (in seconds) as $x = 0.01 \cos \left( \pi t + \frac{\pi}{4} \right)$. The frequency of the motion will be

The figure shows the $x-t$ plot of a particle executing one-dimensional simple harmonic motion. Determine the signs of the position,velocity,and acceleration variables of the particle at $t = 0.3 \; s$,$t = 1.2 \; s$,and $t = -1.2 \; s$.

$A$ particle moves in $S.H.M.$ such that its acceleration is $a = -px$,where '$x$' is the displacement of the particle from the equilibrium position and '$p$' is a constant. The period of oscillation is

Mark the wrong statement.

$A$ 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

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 $t = 2 \ s$.