A particle starts oscillating simple harmonically from its mean...

The equation of displacement for a particle executing simple harmonic motion starting from the mean position is given by:

$x = A \sin(\omega t)$

Given $t = \dfrac{T}{12}$ and $\omega = \dfrac{2\pi}{T}$, the displacement is:

x = A \sin\left(\dfrac{2\pi}{T} \times \dfrac{T}{12}\right) = A \sin\left(\dfrac{\pi}{6}\right) = A \sin(30^{}^{\circ}) = \dfrac{A}{2}

The potential energy ($U$) of the particle is:

$U = \dfrac{1}{2}kx^2 = \dfrac{1}{2}k\left(\dfrac{A}{2}\right)^2 = \dfrac{1}{8}kA^2$

The total energy ($E$) of the particle is:

$E = \dfrac{1}{2}kA^2$

The kinetic energy ($K$) is the difference between total energy and potential energy:

$K = E - U = \dfrac{1}{2}kA^2 - \dfrac{1}{8}kA^2 = \dfrac{3}{8}kA^2$

The ratio of potential energy to kinetic energy is:

$\dfrac{U}{K} = \dfrac{\dfrac{1}{8}kA^2}{\dfrac{3}{8}kA^2} = \dfrac{1}{3}$

Thus, the required ratio is $1 : 3$.

Answer: $1 : 3$