$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be

$A$ particle of mass $m$ is executing simple harmonic motion about its mean position. If $A$ is the amplitude and $T$ is the period of $S$.$H$.$M$.,then the total energy of the particle is

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 $\text{ unit}$)

An object of mass $0.2 \ kg$ executes simple harmonic motion along the $X-$ axis with a frequency of $\frac{25}{\pi} \ Hz$. At the position $x = 0.04 \ m$,the object has a kinetic energy of $0.5 \ J$ and a potential energy of $0.4 \ J$. The amplitude of oscillation in meters is equal to:

The displacement of a simple harmonic motion of amplitude $6 \text{ cm}$ when its kinetic energy is equal to its potential energy is:

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t=\frac{T}{12}$,the ratio of the potential energy to kinetic energy of the particle is $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$