Physics - SHM Question with Solution | TestHub

Let distance be$x_1$when velocity is u and acceleration$\alpha .$
Let distance be$x_2$when velocity is v and acceleration$\beta .$
If$\omega$is the angular frequency then
$\alpha = {\omega }^2 x_1$
And$\beta ={\omega }^2 x_2$
$\therefore \alpha +\beta ={\omega }^2(x_1+x_2)$....... (i)
Also$u^2={\omega }^2{A}^2-{\omega }^2{x}_1^2$
And$v^2={\omega }^2 A^2-{\omega }^2 x_2^2$
$v^2-u^2={\omega }^2(x_1^2-x_2^2)$
$v^2-u^2={\omega }^2(x_1-x_2)(x_1+x_2)$........... (ii)
By Equation. (i) we get$v^2-u^2=(x_1-x_2)(\alpha +\beta )$
$\therefore x_1-x_2=\frac{v^2-u^2}{\alpha +\beta }$
Or$x_2-x_1=\frac{u^2-v^2}{\alpha +\beta }$