At some instant, velocity and acceleration of a particle of mass ‘m’ in SHM are ‘a’ and ‘b’ respectively. If its angular frequency is $\omega$  , then its maximum K.E is

Given that v = a, a = b Acceleration (a) = b

${\text{a}}^{\text{2}}={\omega }^2\left({\text{A}}^{\text{2}}-y^2\right)\cdots \cdots \left(1\right)$

But $\text{a}={\omega }^{2}y;\text{b}={\omega }^{2}y;{\text{b}}^{\text{2}}={\omega }^4{y}^2$

$\frac{{\text{b}}^{\text{2}}}{{\omega }^2}={\omega }^2{y}^2\cdots \left(2\right)$

From (1) & (2)

${\text{a}}^{\text{2}}+\frac{{\text{b}}^2}{{\omega }^2}={\omega }^2{\text{A}}^{\text{2}}-{\omega }^2{y}^2+{\omega }^2{y}^2={\omega }^2{\text{A}}^{\text{2}}$

${\omega }^2=\frac{\left({\text{a}}^{\text{2}}+\frac{{\text{b}}^{\text{2}}}{{\omega }^2}\right)}{{\text{A}}^{\text{2}}}$

$\text{KE}=\frac{1}{2}\text{m}{\omega }^2{\text{A}}^{\text{2}}$

$\text{KE}=\frac{1}{2}\text{m}\frac{\left({\text{a}}^{\text{2}}+\frac{{\text{b}}^{\text{2}}}{{\omega }^2}\right)}{{\text{A}}^{\text{2}}}$

$\text{KE}=\frac{1}{2}\text{m}\left({\text{a}}^{\text{2}}+\frac{{\text{b}}^{\text{2}}}{{\omega }^2}\right)$