$A$ particle is performing $SHM$ with a path length of $8 \ cm$. If it takes $1 \ s$ to reach from the mean position to the extreme position,what is its velocity at the equilibrium position?

$A$ particle is performing $S.H.M.$ with a maximum velocity $V$. If the amplitude is doubled and the periodic time is reduced to $\left(\frac{1}{3}\right)^{\text{rd}}$ of its original value,then the new maximum velocity is:

When a particle executes simple harmonic motion,the nature of the graph of velocity as a function of displacement will be:

$A$ particle executing simple harmonic motion starts from the mean position with amplitude $A$ and periodic time $T$. At what displacement is its speed one-fourth of the maximum speed?

$A$ particle performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period will the velocity of the particle be equal to half of its maximum velocity?

$A$ particle performing $SHM$ is found at its equilibrium position at $t = 1 \, s$. It is found to have a speed of $0.25 \, m/s$ at $t = 2 \, s$. If the period of oscillation is $6 \, s$,calculate the amplitude of oscillation.