Two particles executing $S.H.M.$ of same frequency,meet at $x=+A/2$,while moving in opposite directions. Phase difference between the particles is .........

$A$ 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 $cm/s$).

The displacement of a simple harmonic oscillator after $3 \; s$ starting from its mean position is equal to half of its amplitude. The time period of the harmonic motion is $\dots \; s$.

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 $^o$)

$A$ 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:

$A$ 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?