$A$ body is executing simple harmonic motion with an angular frequency $2\,rad/s$. The velocity of the body at $20\,mm$ displacement,when the amplitude of motion is $60\,mm$,is ...... $mm/s$.

$A$ particle is performing $S.H.M.$ with an amplitude $4 \,cm$. At the mean position, the velocity of the particle is $12 \,cm/s$. When the speed of the particle becomes $6 \,cm/s$, the distance of the particle from the mean position is:

The maximum speed of a particle executing $S.H.M.$ is $1 \ m/s$ and its maximum acceleration is $1.57 \ m/s^2$. The time period of the particle will be .... $s$.

The velocity of a particle executing simple harmonic motion along the $x$-axis is described by the equation $v^2 = 50 - x^2$,where $x$ represents displacement. If the time period of the motion is $\frac{x}{7} \ \text{s}$,the value of $x$ is . . . . . . .

$A$ particle executes $S.H.M.$ with amplitude $a$ and time period $T$. The displacement of the particle when its speed is half of its maximum speed is $\frac{\sqrt{x} a}{2}$. The value of $x$ is $\ldots \ldots \ldots$

If displacement $x$ and velocity $v$ are related as $4v^2 = 16 - x^2$ in a $SHM$,then the time period of the given $SHM$ is (consider $SI$ units).