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 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 $\sqrt{\alpha} \ cm$,where $\alpha = $ . . . . . . .

$A$ particle executes $S.H.M.$ starting from the mean position. Its amplitude is '$a$' and its periodic time is '$T$'. At a certain instant,its speed '$u$' is half that of maximum speed $V_{\text{max}}$. The displacement of the particle at that instant is

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

Obtain the instantaneous velocity of a particle executing $SHM$.

$A$ simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be