$A$ particle executes simple harmonic motion with a periodic time of $0.05 \, s$ and an amplitude of $4 \, cm$. What is its maximum velocity?

$A$ particle is performing $S.H.M.$ about $x = 0$, with an amplitude $a$ and time period $T$. The speed of the particle at $x = \frac{a}{3}$ will be

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

The velocity of a particle executing $S.H.M.$ varies with displacement $(x)$ as $4V^2 = 50 - x^2$. The time period of oscillation is $\frac{x}{7}$ seconds. The value of '$x$' is (Take $\pi = \frac{22}{7}$)

$A$ particle executes $S.H.M.$ of period $\frac{2 \pi}{\sqrt{3}} \text{ s}$ along a straight line $4 \text{ cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is (in $\text{ cm}$)

In case of a simple harmonic motion, if the velocity is plotted along the $X$-axis and the displacement (from the equilibrium position) is plotted along the $Y$-axis, the resultant curve is an ellipse with the ratio:
Major axis (along $X$) $= 20 \pi \times$ Minor axis (along $Y$).
What is the frequency of the simple harmonic motion?