A particle is executing SHM along a straight line. Its velocities at distances X1 and X2 from mean position are V1 and V2, respecilvely. Its time period is

V1=ωA2−x12⇒V12ω2=A2−x12⇒A2=V12ω2+x12V2=ωA2−x22⇒V22ω2=A2−x22⇒A2=V22ω2+x22V12ω2+x12=V22ω2+x22⇒V12−V22ω2=x22−x12ω2=V12−V22x22−x12⇒T=2πx22−x12V12−V22

A particle is executing SHM along a straight line. Its velocities at distances X1 and X2 from mean position are V1 and V2, respecilvely. Its time period is

V1=ωA2−x12⇒V12ω2=A2−x12⇒A2=V12ω2+x12V2=ωA2−x22⇒V22ω2=A2−x22⇒A2=V22ω2+x22V12ω2+x12=V22ω2+x22⇒V12−V22ω2=x22−x12ω2=V12−V22x22−x12⇒T=2πx22−x12V12−V22

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A particle is executing SHM along a straight line. Its velocities at distances X₁ and X₂ from mean position are V₁and V₂, respecilvely. Its time period is

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$2 π \sqrt{\frac{V_{1}^{2} + V_{2}^{2}}{X_{1}^{2} + X_{2}^{2}}}$

$2 π \sqrt{\frac{V_{1}^{2} - V_{2}^{2}}{X_{1}^{2} - X_{2}^{2}}}$

$2 π \sqrt{\frac{X_{1}^{2} + X_{2}^{2}}{V_{1}^{2} + V_{2}^{2}}}$

$2 π \sqrt{\frac{X_{2}^{2} - X_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$

answer is D.

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Detailed Solution

$\begin{array}{l} V_{1} = ω \sqrt{A^{2} - x_{1}^{2}} \\ \Rightarrow \frac{V_{1}^{2}}{ω^{2}} = A^{2} - x_{1}^{2} \Rightarrow A^{2} = \frac{V_{1}^{2}}{ω^{2}} + x_{1}^{2} \\ V_{2} = ω \sqrt{A^{2} - x_{2}^{2}} \\ \Rightarrow \frac{V_{2}^{2}}{ω^{2}} = A^{2} - x_{2}^{2} \Rightarrow A^{2} = \frac{V_{2}^{2}}{ω^{2}} + x_{2}^{2} \\ \frac{V_{1}^{2}}{ω^{2}} + x_{1}^{2} = \frac{V_{2}^{2}}{ω^{2}} + x_{2}^{2} \Rightarrow \frac{V_{1}^{2} - V_{2}^{2}}{ω^{2}} = x_{2}^{2} - x_{1}^{2} \\ ω^{2} = \frac{V_{1}^{2} - V_{2}^{2}}{x_{2}^{2} - x_{1}^{2}} \Rightarrow T = 2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}} \end{array}$