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 ..... $rad/s$.

Two $S$.$H$.Ms. are represented by equations $y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} \right)$ and $y_2 = 0.1 \cos (100 \pi t)$. The phase difference between the speeds of the two particles is

$A$ 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: $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$).

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

What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is