A particle is making simple harmonic motion along the x-axis. If at a distance x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively, then the time period of its oscillation is given as _____

V1=ωA2−x12V12=ω2A2−x12ω2−−−(1),V12=ω2A2−x22ω2−−(2)From (1) and (2) ω2=V12−V22x22−x12⇒T=2πx22−x12V12−V22

A particle is making simple harmonic motion along the x-axis. If at a distance x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively, then the time period of its oscillation is given as _____

V1=ωA2−x12V12=ω2A2−x12ω2−−−(1),V12=ω2A2−x22ω2−−(2)From (1) and (2) ω2=V12−V22x22−x12⇒T=2πx22−x12V12−V22

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A particle is making simple harmonic motion along the x-axis. If at a distance x₁ and x₂ from the mean position the velocities of the particle are v₁ and v₂ respectively, then the time period of its oscillation is given as _____

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

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

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

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

answer is D.

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Detailed Solution

$V_{1} = ω \sqrt{A^{2} - x_{1}^{2}}$
$V_{1}^{2} = ω^{2} A^{2} - x_{1}^{2} ω^{2} - - - (1), V_{1}^{2} = ω^{2} A^{2} - x_{2}^{2} ω^{2} - - (2)$
From (1) and (2) $ω^{2} = \frac{V_{1}^{2} - V_{2}^{2}}{x_{2}^{2} - x_{1}^{2}} \Rightarrow T = 2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$