Physics - SHM Question with Solution | TestHub

Let the particle perform S.H.M. with angular frequency $\omega$ and amplitude $A$.

The acceleration of the particle at displacement $x$ from the mean position is given by $a=-{\omega }^{2}x$.

Given that the accelerations at two positions are $\alpha$ and $\beta$ with $0<\alpha <\beta$, both accelerations are in the same direction. Thus, the two positions lie on the same side of the mean position.

Let the displacements be $x_1$ and $x_2$.

$\alpha =-{\omega }^2{x}_1\Rightarrow x_1=-\frac{\alpha }{{\omega }^2}$

$\beta =-{\omega }^2{x}_2\Rightarrow x_2=-\frac{\beta }{{\omega }^2}$

The distance between the two positions is $d=|x_2-x_1|=\frac{\beta -\alpha }{{\omega }^2}$.

The velocities at these positions are $u$ and $v$, given by:

$u^2={\omega }^2(A^2-x_1^2)$

$v^2={\omega }^2(A^2-x_2^2)$

Subtracting the two equations, we get:

$u^2-v^2={\omega }^2(x_2^2-x_1^2)$

Substituting $x_1$ and $x_2$:

$u^2-v^2={\omega }^2\left(\frac{{\beta }^2}{{\omega }^4}-\frac{{\alpha }^2}{{\omega }^4}\right)$

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

$u^2-v^2=\left(\frac{\beta -\alpha }{{\omega }^2}\right)(\beta +\alpha )$

Since $d=\frac{\beta -\alpha }{{\omega }^2}$, we have:

$u^2-v^2=d(\alpha +\beta )$

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

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