Discussion Forum

A particle performs linear SHM at a particular instant, velocity of the particle is 'u' and acceleration is '$\alpha$' while at another instant velocity is 'v' and acceleration is '$\beta$  ( 0 <$\alpha$ < $\beta$ ). The distance between the two positions is 

Answer:

Explanation:

Let  the distance be p when velocity is u and acceleration $\alpha$.

 Let the distance  q when velocity is v and acceleration $\beta$.

 If $\omega$  is the angular frequency  , then 

$\alpha=\omega^{2} p$   ans $\beta=\omega^{2} q$

 $\therefore$      $\alpha +\beta$= $\omega ^{2} (p+q)$      ........(i)

 Also,    $u^{2}=\omega ^{2} A^{2}-\omega ^{2} p^{2}$

and    $v^{2}=\omega ^{2} A^{2}-\omega ^{2} q^{2}$

 $\Rightarrow$   $v^{2}-u^{2}= \omega ^{2}(p^{2}-q^{2})$

 $v^{2}-u^{2}= \omega ^{2}(p^{}-q^{})(p+q)$    .........(ii)

 by Eqs .(i) and (ii), we get

$v^{2}-u^{2}= (p^{}-q^{})(\alpha+\beta)$

 $\therefore$   $p-q= \frac{v^{2}-u^{2}}{\alpha+\beta}$  or   $q-p= \frac{u^{2}-v^{2}}{\alpha+\beta}$