After what minimum time does a particle attain \[KE = \dfrac{1}{3}PE\] given that the time period \[T\] and initially, the particle is at the positive extreme position?

Hint:We start by noting down the data given in the question. We then get a relation between the amplitude and the distance at which the given condition is valid. We compare the relationship with the standard equation for position to get a value of phase. Once the value of phase at this point is acquired, we apply this to the relation between time and phase to get the minimum time.

Formulas used:
The kinetic energy at a point between the mean and extreme position is,
\[KE = \dfrac{1}{2}k\left( {{A^2} - {x^2}} \right)\]
The potential energy at a point between the mean and extreme position is,
\[PE = \dfrac{1}{2}k{x^2}\]
Where, \[x\] is the distance at which the value of kinetic energy is one third of that potential energy and \[A\] is the amplitude of the system.

Complete step by step answer:
The following information is given: the relation between kinetic energy and potential energy is, \[KE = \dfrac{1}{3}PE\]. The time period of this motion is, \[T\] and the particle is initially at the positive extreme position. Now we compute the distance at which the value of kinetic energy is one third that of potential energy.
\[KE = \dfrac{1}{3}PE\]

We substitute the values and get
\[\dfrac{1}{3} \times \dfrac{1}{2}k{x^2} = \dfrac{1}{2}k\left( {{A^2} - {x^2}} \right)\]
Cancelling the same values from both the sides, we arrive at,
\[\dfrac{1}{3}{x^2} = \left( {{A^2} - {x^2}} \right)\]
Now we cross multiply the three to the other side and bring the like terms to one side.
\[4{x^2} = 3{A^2}\]
Now we take the square root and bring the known values to one side to get,
\[x = \dfrac{{\sqrt 3 }}{2}A\]
This means that the phase covered is, \[\dfrac{\pi }{6}\]

Now that we have the value of phase, we can find the value of time using the formula,
\[t = phase \times period \times \dfrac{1}{{2\pi }}\]
The two pi is used to convert radians,
The time is hence,
\[t = phase \times period \times \dfrac{1}{{2\pi }} \\
\Rightarrow t= \dfrac{\pi }{6} \times T \times \dfrac{1}{{2\pi }} \\
\therefore t= \dfrac{T}{{12}}\]

Hence, the minimum time after that particle is at the positive extreme position is $\dfrac{T}{{12}}$.

Note:Time period is defined as the time taken for one complete revolution or vibration. Hence, the time taken for extreme to mean position will be lesser than the actual period. The time period can also be defined as the reciprocal of the frequency. This relation is taken into account in some problems as it makes the calculations easier.