The displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is:

Two particles are oscillating along two close parallel straight lines side by side,with the same frequency and amplitudes. They pass each other,moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

$A$ simple harmonic oscillator of frequency $1 \ Hz$ has a phase of $1$ radian. By how much should the origin be shifted in time so as to make the phase of the oscillator vanish? ($t$ in seconds).

$A$ particle executes $SHM$ of amplitude $25\, cm$ and time period $3\, s$. What is the minimum time required for the particle to move between two points $12.5\, cm$ on either side of the mean position?

The displacement-time equation of a particle executing $SHM$ is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$. The time taken by the particle to go directly from $x = -\frac{A}{2}$ to $x = +\frac{A}{2}$ is