A particle executes SHM on a straight line- At two positions it-s velocity u and v while acceleration- - and - respectively - - - -0 - the distance between the these two positions is

The correct option is B u ^ 2 v ^ 2 α +β u=w√( A ^ 2 x _ 1 ^ 2 ) v=w√( A ^ 2 x _ 2 ^ 2 ) α = w ^ 2 x _ 1 β = w ^ 2 x _ 2 ...

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Standard XII

Physics

Phase and the General Equation

A particle ex...

Question

A particle executes SHM on a straight line. At two positions it's velocity $u$ and $v$ while acceleration, $α$ and $β$ respectively $[β > α > 0] .$ the distance between the these two positions is

\frac{u^{2} + v^{2}}{α + β}

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\frac{u^{2} - v^{2}}{α + β}

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\frac{u^{2} - v^{2}}{β - α}

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\frac{u^{2} + v^{2}}{β - α}

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Solution

The correct option is B $\frac{u^{2} - v^{2}}{α + β}$
$u = w \sqrt{A^{2} - x_{1}^{2}} v = w \sqrt{A^{2} - x_{2}^{2}} α = - w^{2} x_{1} β = - w^{2} x_{2} \Rightarrow - (x_{1} + x_{2}) = \frac{d}{w^{2}} (α + β) \frac{u^{2} - v^{2}}{w^{2}} = - (x_{1}^{2} - x_{2}^{2}) = - (x_{1} + x_{2}) (x_{1} - x_{2}) = \frac{(α + β)}{w^{2}} (x_{1} - x_{2}) \Rightarrow x_{1} - x_{2} = \frac{u^{2} - v^{2}}{α + β}$

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A particle executes SHM on a straight line. At two positions it's velocity u and v while acceleration, α and β respectively [β>α>0]. the distance between the these two positions is

The correct option is B u2−v2α+βu=w√A2−x21v=w√A2−x22α=−w2x1β=−w2x2⇒−(x1+x2)=dw2(α+β)u2−v2w2=−(x21−x22)=−(x1+x2)(x1−x2)=(α+β)w2(x1−x2)⇒x1−x2=u2−v2α+β

A particle executes SHM on a straight line. At two positions it's velocity $u$ and $v$ while acceleration, $α$ and $β$ respectively $[β > α > 0] .$ the distance between the these two positions is