A particle is executing simple harmonic motion along a straight line. At displacements $r_{1}$ and $r_{2}$ from its mean position the velocities are $v_{1}$ and $v_{2}$ . The time period of the particle is:

{2 π [\frac{v_{1}^{2} - v_{2}^{2}}{r_{2}^{2} + r_{1}^{2}}]}^{1 / 2}

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{\frac{1}{2 π} [\frac{v_{1}^{2} + v_{2}^{2}}{r_{2}^{2} - r_{1}^{2}}]}^{1 / 2}

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{2 π [\frac{r_{2}^{2} - r_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}]}^{1 / 2}

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{2 π [\frac{r_{2}^{2} - r_{1}^{2}}{v_{2}^{2} - v_{1}^{2}}]}^{1 / 2}

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Solution

The correct option is D ${2 π [\frac{r_{2}^{2} - r_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}]}^{1 / 2}$
Let the SHM be represented by $x = A s i n ω t$
Thus $v = A ω c o s ω t$
$⟹ r_{1} = A s i n ω t_{1}$
$v_{1} = A ω c o s ω t_{1}$
$⟹ r_{1}^{2} + \frac{v_{1}^{2}}{ω^{2}} = A^{2}$
Similarly, $r_{2}^{2} + \frac{v_{2}^{2}}{ω^{2}} = A^{2}$
$⟹ r_{1}^{2} + \frac{v_{1}^{2}}{ω^{2}} = r_{2}^{2} + \frac{v_{2}^{2}}{ω^{2}}$
$⟹ ω = \sqrt{\frac{v_{1}^{2} - v_{2}^{2}}{r_{2}^{2} - r_{1}^{2}}}$
Thus time period of particle= $T = \frac{2 π}{ω}$
$= 2 π [\frac{r_{2}^{2} - r_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}]^{1 / 2}$

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