The kinetic energy of a simple harmonic oscillator is oscillating with an angular frequency of $176 \ rad/s$. The frequency of this simple harmonic oscillator is . . . . . . $Hz$. $\left[\text{take } \pi=\frac{22}{7}\right]$

$A$ particle executing simple harmonic motion starts from the mean position with amplitude $A$ and periodic time $T$. At what displacement is its speed one-fourth of the maximum speed?

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

$A$ 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:

The maximum velocity of a simple harmonic motion represented by $y = 3\sin \left( 100t + \frac{\pi}{6} \right)$ is given by

$A$ spring executes $SHM$ with a mass of $10\,kg$ attached to it. The force constant of the spring is $10\,N/m$. If at any instant its velocity is $40\,cm/s$,the displacement will be .... $m$ (where amplitude is $0.5\,m$).