After the beginning of motion,how long will it take for a harmonically oscillating particle to reach a displacement equal to one-half of its amplitude,if the time period is $24 \ sec$ and the particle starts from rest?

$A$ large number of random snapshots using a camera are taken of a particle in simple harmonic motion between $x = -x_0$ and $x = +x_0$,with the origin $x = 0$ as the mean position. $A$ histogram of the total number of times the particle is recorded about a given position (Event no.) would most closely resemble:

$A$ particle executing $S.H.M.$ with an amplitude of $4 \, cm$ and a time period $T = 4 \, s$. The time taken by it to move from the positive extreme position to half the amplitude is ..... $s$.

The displacement-time equation of a particle executing $SHM$ is $x = A \sin(\omega t + \phi)$. At time $t = 0$,the position of the particle is $x = A/2$ and it is moving along the negative $x$-direction. Then the phase angle $\phi$ is:

The displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.

$A$ particle executes $SHM$ with an amplitude of $20 \, cm$ and a time period of $12 \, s$. What is the minimum time required for it to move between two points $10 \, cm$ on either side of the mean position?