## College Physics (4th Edition)

$v_m^2 = a_m~A$
We can write an expression for $v_m^2$: $v_m^2 = (A~\omega)^2 = A^2~\omega^2$ We can write an expression for $a_m$: $a_m = A~\omega^2$ Then: $v_m^2 = A^2~\omega^2 = (A~\omega^2)~A = a_m~A$