$A$ simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be

$A$ particle executes linear $S.H.M.$ with amplitude $A = 4 \text{ cm}$. The magnitude of velocity and acceleration is equal when it is at a distance $x = 3 \text{ cm}$ from the mean position. The time period of the oscillation is:

$A$ particle executes simple harmonic motion with a periodic time of $0.05 \, s$ and an amplitude of $4 \, cm$. What is its maximum velocity?

$A$ particle is executing simple harmonic motion with amplitude $A$. At a distance $x$ from the mean position,when the particle is moving towards the extreme position,it receives a blow in the direction of motion which instantaneously doubles its velocity. What is the new amplitude of the particle? (Frequency remains constant during the motion)

The displacements of two particles executing simple harmonic motion are represented as $y_{1} = 2 \sin (10 t + \theta)$ and $y_{2} = 3 \cos 10 t$. The phase difference between the velocities of these waves is

The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is