A particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium is at $x = 0,$ force constant is $K$ and amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be
$\frac{{{K^{\frac{3}{2}}}{A^2}}}{{\sqrt m }}$
$\frac{{2{K^{\frac{3}{2}}}{A^2}}}{{\sqrt m }}$
$\frac{{{K^{\frac{3}{2}}}{A^2}}}{{3\sqrt m }}$
$\frac{{{K^{\frac{3}{2}}}{A^2}}}{{2\sqrt m }}$
When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y ( t )= y _{0} \sin ^{2} \omega t ,$ where $'y'$ is measured from the lower end of unstretched spring. Then $\omega$ is
Two springs having spring constant $k_1$ and $k_2$ is connected in series, its resultant spring constant will be $2\,unit$. Now if they connected in parallel its resultant spring constant will be $9\,unit$, then find the value of $k_1$ and $k_2$.
In the following questions, match column $-I$ with column $-II$ and choose the correct options
If a spring extends by $x$ on loading, then energy stored by the spring is (if $T$ is the tension in the spring and $K$ is the spring constant)
The time period of simple harmonic motion of mass $\mathrm{M}$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5 K}}$, where the value of $\alpha$ is____.