$A$ particle starts $S.H.M.$ from the mean position as shown in the figure. Its amplitude is $A$ and its time period is $T$. At a certain instant,its speed is half of its maximum speed. What is its displacement at this instant?

$A$ 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?

$A$ body is executing $S.H.M.$ When its displacement from the mean position is $4 \, cm$ and $5 \, cm$,the corresponding velocity of the body is $10 \, cm/sec$ and $8 \, cm/sec$. Then the time period of the body is

$A$ point mass of $400 \, g$ executes $S.H.M.$ under a force $F = -(10 \, N m^{-1}) x$. If it crosses the centre of oscillation with a speed of $10 \, m s^{-1}$, the amplitude of motion is (in $ \, m$)

$A$ 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

What is the maximum velocity of a simple harmonic motion with an amplitude of $2 \ m$ and a time period of $2 \ s$?