$A$ particle is executing $SHM$ with an amplitude $A$. What is the displacement of the particle when its potential energy is half of its total energy?

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

$A$ body is executing $S$.$H$.$M$. Its potential energy is '$P_1$' and '$P_2$' at displacements '$x$' and '$y$' respectively. The potential energy at displacement $(x+y)$ is

$A$ particle executing simple harmonic motion along a straight line with an amplitude $A$,attains maximum potential energy when its displacement from the mean position equals

$A$ body is executing $S.H.M$. At a displacement $x$ its potential energy is $9 \ J$ and at a displacement $y$ its potential energy is $16 \ J$. The potential energy at displacement $(x+y)$ is (in $J$)

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: