Question

The energy of a particle executing simple harmonic motion is given by the equation $E=A{x}^2+B{v}^2$ where $x$ is the displacement from mean position $x=0$ and $v$ is the velocity of the particle at $x$. Find the amplitude of $S.H.M$.

Answer 1

$E=A{x}^2+B{v}^2$

As we know that at max amplitude, V$=0$

$\therefore E=A{a}^2$

Let a be amplitude

$\therefore a^2=\frac{E}{A}$

$|a|=\sqrt{\frac{E}{A_1}}$.