In $S.H.M.$,the displacement of a particle at an instant is $Y = A \cos 30^{\circ}$,where $A = 40 \ cm$ and kinetic energy is $200 \ J$. If the force constant is $1 \times 10^{x} \ N/m$,then $x$ will be $(\cos 30^{\circ} = \sqrt{3}/2)$.

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $\omega_1$ and $\omega_2$ and have total energies $E_1$ and $E_2$,respectively. The variations of their momenta $p$ with positions $x$ are shown in the figures. If $\frac{a}{b}= n^2$ and $\frac{a}{R}= n$,then the correct equation$(s)$ is(are):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)

Vibrational motion possesses which type of energy?

Which graph represents the difference between total energy and potential energy of a particle executing $SHM$ versus its distance from the mean position?

$A$ particle of mass $m$ is executing $S.H.M.$ about the origin on the $x$-axis with frequency $f = \frac{\sqrt{Ka}}{\pi m}$, where $K$ is a constant and $a$ is the amplitude of $S.H.M.$ If $x$ is the displacement of the particle at time $t$, the potential energy of the particle will be: