If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \text{ m}$ represents the motion of a particle executing simple harmonic motion,the amplitude and time period of motion,respectively,are

$A$ particle moves with simple harmonic motion in a straight line. In the first $\tau \ s$,after starting from rest,it travels a distance $a$,and in the next $\tau \ s$ it travels $2a$ in the same direction. Then:

$A$ mass $m = 100 \, g$ is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to $0.16 \, m$ and a time period equal to $2 \, s$. Initially,the mass is released from rest at $t = 0$ and displacement $x = -0.16 \, m$. The expression for the displacement of the mass at any time $t$ is:

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 $\pi = \frac{22}{7}$):

The equation of motion of a particle is $x = a \cos(\alpha t)$. The motion is

Fill in the blanks:
$1.$ The ratio of displacement at any position and ....... remains constant for a particle executing $SHM$.
$2.$ The radius of the reference circle is equal to the .......... of the oscillator.
$3.$ Increase in phase per second of $SHO =$ ......... .
$4.$ $SHO$ covers ......... distance in one periodic time.