Question

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

• A. $\frac{u^2-v^2}{\alpha +\beta }$
• B. $\frac{u^2+v^2}{\alpha +\beta }$
• C. $\frac{u^2-v^2}{\alpha -\beta }$
• D. $\frac{u^2+v^2}{\alpha -\beta }$

Answer 1

Answer: (A)

$\frac{u^2-v^2}{\alpha +\beta }$

$u=\omega \sqrt{(A^2-x^2)}$

$\alpha =-{\omega }^{2}x$

$v=\omega \sqrt{(A^2-y^2)}$

$\beta =-{\omega }^{2}y$

on squaring and subtracting the velocity expression

we get $u^2-v^2={\omega }^2(x+y)(y-x)$ putting the value of $alpha$ and $\beta$

we get $x-y=\frac{u^2-v^2}{\alpha +\beta }$