For a simple harmonic motion given by the equation $x = 5 \sin (4t - \frac{\pi}{6}) \, m$,what is the velocity when its displacement is $3 \, m$ (in $, m/s$)?

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $16 \ s$. After $2 \ s$,its velocity is $\pi \ m \ s^{-1}$. The amplitude of oscillation is (Given: $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$).

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

$A$ spring executes $S.H.M.$ with a mass of $1 \,kg$ attached to it. The force constant of the spring is $4 \,N/m$. If at any instant its velocity is $20 \,cm/s$, the displacement at that instant is (Amplitude of $S.H.M.$ is $0.4 \,m$)

$A$ particle executing $S.H.M.$ has velocities $V_1$ and $V_2$ at distances $x_1$ and $x_2$ respectively,from the mean position. Its frequency is

$A$ particle executing simple harmonic motion has a maximum speed of $40 \,ms^{-1}$ and maximum acceleration of $60 \,ms^{-2}$. The period of oscillation is