What is the maximum velocity of a simple harmonic motion with an amplitude of $3 \ cm$ and a time period of $6 \ s$?

The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is

$A$ particle executes a linear $S.H.M.$ In two of its positions,the velocities are $V_1$ and $V_2$,and the accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$. The distance between the positions is

The maximum velocity of a particle,executing simple harmonic motion with an amplitude $7 \ mm$,is $4.4 \ m/s$. The period of oscillation is .... $sec$

$A$ 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 . . . . . . .

The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are $2 \ m/s$ and $4 \ m/s^2$,respectively. Then the angular velocity will be ..... $rad/s$.