Physics - SHM Question with Solution | TestHub

$v^2={\omega }^2\left(A^2-x^2\right)$

$\begin{array}{rlrl}& a=-{\omega }^{2}x& & \end{array}$

$v_1^2={\omega }^2\left[A^2-{\left(\frac{a_1}{{\omega }^2}\right)}^2\right]\dots \left(1\right)$

$v_2^2={\omega }^2\left[A^2-{\left(\frac{a_2}{{\omega }^2}\right)}^2\right]\dots \left(2\right)$

From equations (1) and (2)

$\begin{array}{rlr}& \frac{v_1^2}{{\omega }^2}+\frac{a_1^2}{{\omega }^4}=\frac{v_2^2}{{\omega }^2}+\frac{a_2^2}{{\omega }^4}& \end{array}$

$\Rightarrow \omega =\sqrt{\frac{a_2^2-a_1^2}{v_1^2-v_2^2}}$

Substituting w in equation (1),

$A=\sqrt{\frac{v_1^2-v_2^2}{a_2^2-a_1^2}\left[\frac{a_2^2{v}_1^2-a_1^2{v}_2^2}{a_2^2-a_1^2}\right]}\text{ or }\sqrt{\frac{\left(v_1^2-v_2^2\right)}{{\left(a_2^2-a_1^2\right)}^2}\left(a_2^2{v}_1^2-a_1^2{v}_2^2\right)}$