Simple Harmonic Motion
3.0 Solution of the Equation of SHM
3.0 Solution of the Equation of SHM
Equation of SHM is given by,
$$\begin{equation} \begin{aligned} F = - kx \\ ma = - kx \\ \frac{{{d^2}x}}{{d{t^2}}} = - \frac{k}{m}x \\\end{aligned} \end{equation} $$
Multiplying, $\frac{{dx}}{{dt}}$ on both side, $$\frac{{{d^2}x}}{{d{t^2}}}.\frac{{dx}}{{dt}} = - \frac{k}{m}x.\frac{{dx}}{{dt}}$$
On, integrating we get, $$\frac{1}{2}{\left( {\frac{{dx}}{{dt}}} \right)^2} = - \frac{k}{m}\frac{{{x^2}}}{2} + C$$
For, the value of constant $C$, $\frac{{dx}}{{dt}} = v = 0$ at $x = A$ (at extreme positions).
$$\begin{equation} \begin{aligned} C = \frac{k}{m}\frac{{{A^2}}}{2} \\ \frac{1}{2}{\left( {\frac{{dx}}{{dt}}} \right)^2} = - \frac{k}{m}\frac{{{x^2}}}{2} + \frac{k}{m}\frac{{{A^2}}}{2} \\ \frac{1}{2}{\left( {\frac{{dx}}{{dt}}} \right)^2} = \frac{{\left( {{A^2} - {x^2}} \right)}}{2}\frac{k}{m} \\ {\left( {\frac{{dx}}{{dt}}} \right)^2} = \left( {{A^2} - {x^2}} \right)\frac{k}{m} \\ \frac{{dx}}{{dt}} = \omega \sqrt {\left( {{A^2} - {x^2}} \right)} \quad \left( {\because \omega = \sqrt {\frac{k}{m}} } \right) \\ v = \omega \sqrt {\left( {{A^2} - {x^2}} \right)} \\ \frac{{dx}}{{\sqrt {\left( {{A^2} - {x^2}} \right)} }} = \omega dt \\\end{aligned} \end{equation} $$
Integrating the above equation with respect to ‘$t$’, we get,
$$\begin{equation} \begin{aligned} x = A\sin k\;\left( {{\text{where }}k = {\text{constant}}} \right) \\ {\text{or, }}k = {\sin ^{ - 1}}\frac{x}{A} \\ \frac{{A\cos kdk}}{{A\cos k}} = \omega dt \\ dk = \omega dt \\\end{aligned} \end{equation} $$
On integration, $$\begin{equation} \begin{aligned} k = \omega t \\ {\sin ^{ - 1}}\frac{x}{A} = wt + \phi \\ x = A\sin \left( {\omega t + \phi } \right) \\\end{aligned} \end{equation} $$ where, $\phi$ is another constant of integration which depends on the initial condition.
