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Physics Question on Oscillations

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

A

u2−v2α+β\frac{u^2 - v^2}{\alpha + \beta}

B

u2+v2α+β\frac{u^2 + v^2}{\alpha + \beta}

C

u2−v2α−β\frac{u^2 - v^2}{\alpha - \beta}

D

u2+v2α−β\frac{u^2 + v^2}{\alpha - \beta}

Answer

u2−v2α+β\frac{u^2 - v^2}{\alpha + \beta}

Explanation

Solution

Answer (a) u2−v2α+β\frac{u^2 - v^2}{\alpha + \beta}