A particle performs linear S-H-M- At a particular instant- velocity of the particle is -u- and acceleration is - while at another instant velocity is -v- and acceleration is - 0 - - - - The distance between the two positions is

The correct option is B u^2 v^2α + β u=ω√((A^2 x^2)) α= ω^2x v=ω√((A^2 y^2)) β= ω^2y on squaring and subtracting the velocity expres ...

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Byju's Answer

Standard XII

Physics

Instantaneous Velocity

A particle pe...

Question

A particle performs linear S.H.M. At a particular instant, velocity of the particle is $^{'} u^{'}$ and acceleration is $^{'} α^{'}$ while at another instant velocity is $^{'} v^{'}$ and acceleration is $^{'} β^{'} (0 < α < β)$ . The distance between the 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}}{α - β}

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\frac{u^{2} + v^{2}}{α - β}

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Solution

The correct option is B $\frac{u^{2} - v^{2}}{α + β}$
$u = ω \sqrt{(A^{2} - x^{2})}$
$α = - ω^{2} x$
$v = ω \sqrt{(A^{2} - y^{2})}$
$β = - ω^{2} y$
on squaring and subtracting the velocity expression
we get $u^{2} - v^{2} = ω^{2} (x + y) (y - x)$ putting the value of $a l p h a$ and $β$
we get $x - y = \frac{u^{2} - v^{2}}{α + β}$

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A particle performs linear S.H.M. At a particular instant, velocity of the particle is ′u′ and acceleration is ′α′ while at another instant velocity is ′v′ and acceleration is ′β′(0<α<β). The distance between the two positions is

The correct option is B u2−v2α+βu=ω√(A2−x2)α=−ω2xv=ω√(A2−y2)β=−ω2yon squaring and subtracting the velocity expressionwe get u2−v2=ω2(x+y)(y−x) putting the value of alpha and βwe get x−y=u2−v2α+β

A particle performs linear S.H.M. At a particular instant, velocity of the particle is $^{'} u^{'}$ and acceleration is $^{'} α^{'}$ while at another instant velocity is $^{'} v^{'}$ and acceleration is $^{'} β^{'} (0 < α < β)$ . The distance between the two positions is