$A$ simple pendulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=a/2$ will be

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

$A$ particle is executing $SHM$ along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$ respectively. Its time period is

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $16 \ s$. After $2 \ s$,its velocity is $\pi \ m \ s^{-1}$. The amplitude of oscillation is (Given: $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$).

$A$ particle performing $SHM$ has a time period of $\frac{2 \pi}{\sqrt{3}} \,s$ and a path length of $4 \,cm$. The displacement from the mean position at which the magnitude of acceleration is equal to the magnitude of velocity is (in $\,cm$)

For a simple harmonic motion with amplitude $A$ and time period $T$,at what position is the velocity half of the maximum velocity?