The total energy of a particle executing $S.H.M.$ is proportional to

If the amplitude of a body executing $SHM$ is doubled,what will be its energy?

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 amplitude of a particle executing $SHM$ is $3 \text{ cm}$. The displacement at which its kinetic energy will be $25\%$ more than its potential energy is (in $\text{cm}$):

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ 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 $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$