$A$ mass attached to a spring performs $S.H.M.$ whose displacement is $x = 3 \times 10^{-3} \cos(2 \pi t) \text{ m}$. The time taken to obtain maximum speed for the first time is

The displacement of a particle in $S.H.M.$ is given by $x = A \cos(\omega t + \pi/6)$. At what time will its speed be maximum?

Maximum speed of a particle in simple harmonic motion is $v_{max}$. Then,the average speed of a particle in one complete oscillation is equal to:

$A$ particle executes $SHM$. Its velocities are $v_1$ and $v_2$ at displacements $x_1$ and $x_2$ from the mean position,respectively. The frequency of oscillation will be

$A$ particle executing $S.H.M.$ has velocities $V_1$ and $V_2$ at distances $x_1$ and $x_2$ respectively,from the mean position. Its frequency is

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