The velocity at the mean position of a particle executing $S.H.M.$ is $v$. What is the velocity of the particle at a distance equal to half of the amplitude?

$A$ particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

$A$ spring hangs vertically from the ceiling and a mass is attached to its free end. When the mass is pulled down and released,it oscillates vertically with simple harmonic motion of period $T$. The variation with time $t$ of its distance from the ceiling is as shown. Which statement gives a correct deduction from this graph?

$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

$A$ particle executes $S.H.M.$ and its position varies with time as $x = A \sin \omega t$. Its average speed during its motion from mean position to mid-point of mean and extreme position is