A particle is performing simple harmonic motion. If its velocities are $v_{1} and v_{2}$ at the displacements from the mean position arc $y_{1} and y_{2}$ respectively, then its time period is

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$2 π \sqrt{\frac{{v^{2}}_{1} + v^{2}_{2}}{{y^{2}}_{1} + {y^{2}}_{2}}}$

$2 π \sqrt{\frac{{y^{2}}_{1} + {y_{2}}^{2}}{{v_{1}}^{2} + {v_{2}}^{2}}}$

$2 π \sqrt{\frac{{y^{2}}_{1} - {y^{2}}_{2}}{{v_{2}}^{2} - {v^{2}}_{1}}}$

$2 π \sqrt{\frac{{v_{2}}^{2} - {v^{2}}_{1}}{{y^{2}}_{1} - {y^{2}}_{2}}}$

answer is C.

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Detailed Solution

Velocity of a particle in SHM $v = ω \sqrt{A^{2} - y^{2}}$
$v_{1} = ω \sqrt{A^{2} - {y_{1}}^{2}} \Rightarrow {v^{2}}_{1} = ω^{2} A^{2} - ω^{2} {y^{2}}_{1}$ ...(i)
and $v_{2} = ω \sqrt{A^{2} - {y^{2}}_{2}} \Rightarrow {v^{2}}_{2} = ω^{2} A^{2} - ω^{2} {y^{2}}_{2} - - - - - (ii)$
from equations (i) and (ii), we get

${v_{2}}^{2} - {v^{2}}_{1} = ω^{2} ({y_{1}}^{2} - {y^{2}}_{2})$

$ω = \sqrt{\frac{{v^{2}}_{2} - {v^{2}}_{1}}{{y^{2}}_{1} - {y^{2}}_{2}}} \Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{{y^{2}}_{1} - {y^{2}}_{2}}{{v^{2}}_{2} - {v^{2}}_{1}}}$