$A$ 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 $\ldots \ldots \ldots$

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 \times 10^{-2} \cos(\pi t) \text{ m}$. The time at which the maximum speed first occurs is (in $\text{ s}$)

The maximum velocity of a particle,executing $S$.$H$.$M$. with an amplitude $7 \ mm$ is $4.4 \ ms^{-1}$. The period of oscillation is $\left[\pi=\frac{22}{7}\right]$. (in $s$)

$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is . . . . . . .

$A$ particle executes simple harmonic motion with an amplitude of $4 \, cm$. At the mean position,the velocity of the particle is $10 \, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \, cm/s$ is

$A$ particle performs $S.H.M.$ with amplitude $A$. Its speed is tripled at the instant when it is at a distance of $\frac{2A}{3}$ from the mean position. The new amplitude of the motion is