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:

$A$ particle executes simple harmonic motion between $x=-A$ and $x=+A$. If it takes a time $T_1$ to go from $x=0$ to $x=A/2$ and $T_2$ to go from $x=A/2$ to $x=A$,then:

$A$ body is in simple harmonic motion with a time period of $0.5 \ s$ and an amplitude of $1 \ cm$. Find the average velocity in the interval in which it moves from the equilibrium position to half of its amplitude (in $cm/s$).

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 time taken by a particle executing simple harmonic motion of period $T$ to move from the mean position to half the maximum displacement is

$A$ particle executes $SHM$ with amplitude $0.2 \,m$ and time period $24 \,s$. The time required for it to move from the mean position to a point $0.1 \,m$ from the mean position is (in $\,s$)