$A$ particle executes $SHM$ with a period of $1.2 \, s$ and an amplitude of $8 \, cm$. Find the time it takes to travel $3 \, cm$ from the positive extremity of its oscillation.

Two particles are in $SHM$ on the same straight line with amplitudes $A$ and $2A$ and with the same angular frequency $\omega$. It is observed that when the first particle is at a distance $A/\sqrt{2}$ from the origin and moving toward the mean position,the other particle is at the extreme position on the other side of the mean position. Find the phase difference between the two particles. (in $^o$)

The periodic time of a body executing simple harmonic motion is $3 \, s$. After how much interval from time $t = 0$,will its displacement be half of its amplitude?

$A$ particle performs simple harmonic motion with a period of $3 \ s$. The time taken by it to cover a distance equal to half the amplitude from the mean position is $\left[\sin 30^{\circ}=0.5\right]$.

The phase (at a time $t$) of a particle in simple harmonic motion tells:

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: