Obtain the expression of displacement from the force law for simple harmonic motion.

(N/A) The force law for $SHM$ is given by $F = -kx(t)$.
Since $F = ma(t)$ and $k = m\omega^2$,we have $ma(t) = -m\omega^2 x(t)$,which simplifies to $a(t) = -\omega^2 x(t)$.
We know that acceleration $a(t) = \frac{dv}{dt}$ and velocity $v(t) = \frac{dx}{dt}$.
Substituting $a(t) = \frac{dv}{dt}$,we get $\frac{dv}{dt} = -\omega^2 x$.
Using the chain rule,$\frac{dv}{dx} \cdot \frac{dx}{dt} = -\omega^2 x$,which implies $v \frac{dv}{dx} = -\omega^2 x$.
Integrating both sides with respect to $x$: $\int v dv = -\omega^2 \int x dx$.
$\frac{v^2}{2} = -\omega^2 \frac{x^2}{2} + C$.
At extreme position $x = A$,$v = 0$,so $C = \frac{1}{2} \omega^2 A^2$.
Thus,$v^2 = \omega^2 (A^2 - x^2)$,or $v = \pm \omega \sqrt{A^2 - x^2}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{\sqrt{A^2 - x^2}} = \omega dt$.
Integrating both sides: $\sin^{-1}(\frac{x}{A}) = \omega t + \phi$.
Therefore,the expression for displacement is $x(t) = A \sin(\omega t + \phi)$.