A particle is executing SHM along a straight line, its velocities at distances $ {{x}_{1}} $ and $ {{x}_{2}} $ from the mean position are $ {{V}_{1}} $ and $ {{V}_{2}} $, respectively. Its time period is:
 $ A.2\pi \sqrt{\dfrac{V_{1}^{2}+V_{2}^{2}}{x_{1}^{2}+x_{2}^{2}}} $
 $ B.2\pi \sqrt{\dfrac{V_{1}^{2}-V_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}} $
 $ C.2\pi \sqrt{\dfrac{x_{1}^{2}+x_{2}^{2}}{V_{1}^{2}+V_{2}^{2}}} $
 $ D.2\pi \sqrt{\dfrac{x_{2}^{2}-x_{1}^{2}}{V_{1}^{2}-V_{2}^{2}}} $

Hint: In the process to solve the question we have to apply the concept of simple harmonic motion. If a particle moves back and forth about a mean position in such a way that a restoring torque acts on the particle, which is proportional to displacement from mean position, but in the opposite direction from displacement, this motion of the particle is called simple harmonic motion. Relation of simple harmonic motion with velocity of the particle should be applied.

Formula used:
 To solve this problem we have to apply the following relations:-
 $ v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}} $ and $ \omega =\dfrac{2\pi }{T} $.

Complete step-by-step answer:
From the above problem we have the following parameters:-
Velocity of the particle at position, $ {{x}_{1}}={{v}_{1}} $
Velocity of the particle at position, $ {{x}_{2}}={{v}_{2}} $
Let, the amplitude during the SHM be $ A $ and angular velocity be $ \omega $.
Using expression for velocity for SHM we have
 $ v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}} $.
For position, $ {{x}_{1}} $
 $ {{v}_{1}}=\omega \sqrt{{{A}^{2}}-x_{1}^{2}} $ …………….. $ (i) $
Squaring $ (i) $ we get,
 $ v_{1}^{2}={{\omega }^{2}}({{A}^{2}}-x_{1}^{2}) $ ………………. $ (ii) $
For position, $ {{x}_{2}} $
 $ {{v}_{2}}=\omega \sqrt{{{A}^{2}}-x_{2}^{2}} $ ………………… $ (iii) $
Squaring $ (iii) $ we get,
 $ v_{2}^{2}={{\omega }^{2}}({{A}^{2}}-x_{2}^{2}) $ ………………….. $ (iv) $
Subtracting $ (iv) $ from $ (iii) $ we get,
 $ v_{1}^{2}-v_{2}^{2}={{\omega }^{2}}(x_{2}^{2}-x_{1}^{2}) $ ………………. $ (v) $
But we know that $ \omega =\dfrac{2\pi }{T} $ ………………… $ (vi) $ Where $ T $ is the time period
Putting $ (vi) $ in $ (v) $ we get,
 $ v_{1}^{2}-v_{2}^{2}=\dfrac{4{{\pi }^{2}}}{{{T}^{2}}}(x_{2}^{2}-x_{1}^{2}) $
 $ {{T}^{2}}(v_{1}^{2}-v_{2}^{2})=4{{\pi }^{2}}(x_{2}^{2}-x_{1}^{2}) $
 $ {{T}^{2}}=4{{\pi }^{2}}\left( \dfrac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}} \right) $
 $ T=2\pi \sqrt{\dfrac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}} $.

So, the correct answer is “Option D”.

Additional Information: The motion of a body which is repeated after an equal interval of time is called periodic motion. If the body moves to and fro on the same path then the periodic motion is termed as oscillatory motion. Hence, we can say that every oscillatory motion is periodic but every periodic motion is not oscillatory. Example of oscillatory motion is vibrations of prongs of tuning fork.

Note: We should know that simple harmonic motion is the simplest form of oscillatory motion. In fact simple harmonic motion is the type of oscillatory motion which can be expressed as the sine and cosine functions. Therefore, do not use relations of simple harmonic motion in every oscillatory motion. We should use the relation of angular velocity and time period carefully.