For a simple harmonic motion with amplitude $A$ and time period $T$,what is the velocity at $x = \frac{A}{2}$?

$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

$A$ spring executes $S.H.M.$ with mass $10 \ kg$ attached to it. The force constant of the spring is $10 \ N/m$. If at any instant its velocity is $40 \ cm/s$,the displacement at that instant is (Amplitude of $S.H.M.$ $= 0.5 \ m$) (in $m$)

$A$ 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 of mass $5 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$. Its maximum velocity is $100 \ cm/s$. At what distance from the mean position will its velocity be $50 \ cm/s$?

The velocity of a particle executing $SHM$ varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $\frac{x}{7} \ s$. The value of $x$ is $............$ (Take $\pi = \frac{22}{7}$)