Physics - SHM Question with Solution | TestHub

For a particle in linear S.H.M., the acceleration $a$ at position $x$ is given by $a=-{\omega }^{2}x$.

Given $0<a_1<a_2$, both accelerations have the same sign, meaning the particle is on the same side of the mean position.

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

$a_1=-{\omega }^2{x}_1\Rightarrow x_1=-\frac{a_1}{{\omega }^2}$

$a_2=-{\omega }^2{x}_2\Rightarrow x_2=-\frac{a_2}{{\omega }^2}$

The distance between the two positions is $d=|x_1-x_2|=\frac{a_2-a_1}{{\omega }^2}$.

The velocity $v$ at position $x$ is given by $v^2={\omega }^2(A^2-x^2)$.

For the two positions:

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

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

Subtracting the two equations:

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

Substituting $x_1$ and $x_2$:

$u^2-V^2={\omega }^2\left(\frac{a_2^2}{{\omega }^4}-\frac{a_1^2}{{\omega }^4}\right)=\frac{a_2^2-a_1^2}{{\omega }^2}$

$u^2-V^2=\left(\frac{a_2-a_1}{{\omega }^2}\right)(a_1+a_2)$

Since $d=\frac{a_2-a_1}{{\omega }^2}$, we get:

$u^2-V^2=d(a_1+a_2)$

$d=\frac{u^2-V^2}{a_1+a_2}$

Answer: $\frac{{\mathrm{u}}^2-{\mathrm{V}}^2}{{\mathrm{a}}_1+{\mathrm{a}}_2}$