The velocity of a particle in S.H.M. at position $x_{1}$ and $x_{2}$ are $ν_{1}$ and $ν_{2}$ respectively. Determine value of time period and amplitude.

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$T = 2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}};$ $A = \sqrt{\frac{v_{1}^{2} x_{2}^{2} - v_{2}^{2} x_{1}^{2}}{3 (v_{1}^{2} - v_{2}^{2})}}$

$T = 3 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}};$ $A = \sqrt{\frac{2 (v_{1}^{2} x_{2}^{2} - v_{2}^{2} x_{1}^{2})}{v_{1}^{2} - v_{2}^{2}}}$

$T = π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}};$ $A = \frac{v_{1}^{2} x_{2}^{2} - v_{2}^{2} x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}$

$T = 2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}};$ $A = \sqrt{\frac{v_{1}^{2} x_{2}^{2} - v_{2}^{2} x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}}$

answer is D.

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Detailed Solution

$ν = ω \sqrt{A^{2} - x^{2}} \Rightarrow ν^{2} = ω^{2} (A^{2} - x^{2})$

At position $x_{1}$ velocity $v_{1}^{2} = ω^{2} (A^{2} - x_{1}^{2}) . . . (1)$

At position $x_{2}$ velocity $v_{2}^{2} = ω^{2} (A^{2} - x_{2}^{2}) . . . (2)$

Subtracting $(1)$ from $(2)$ $v_{1}^{2} - v_{2}^{2} = ω^{2} (x_{2}^{2} - x_{1}^{2}) \Rightarrow ω = \sqrt{\frac{v_{1}^{2} - v_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}}$

Time period $T = \frac{2 π}{ω} \Rightarrow T = 2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}}$

Dividing $(1)$ from $(2)$ $\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{A^{2} - x_{1}^{2}}{A^{2} - x_{2}^{2}} \Rightarrow v_{1}^{2} A^{2} - v_{1}^{2} x_{2}^{2} = v_{2}^{2} A^{2} - v_{2}^{2} x_{1}^{2}$

So $A^{2} (v_{1}^{2} - v_{2}^{2}) = v_{1}^{2} x_{2}^{2} - v_{2}^{2} x_{1}^{2} \Rightarrow A = \sqrt{\frac{v_{1}^{2} x_{2}^{2} - v_{2}^{2} x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}}$