Foundations - Topic B - Energy - Exercises - Page 24: B.3(a)

The equation of motion of a harmonic oscillator is given by $m\frac{d^2x}{dt^2}=–k_fx$ or, $\frac{d^2x}{dt^2}=–\frac{k_f}{m}x$ or, $\frac{d^2x}{dt^2}=–\omega^2x$ $\bigg($where $\omega=\sqrt {\frac{k_f}{m}}$$\bigg)$ ..........$(1)$ Let $x=x_0e^{kt}$ is the solution of eq, $1$, where $A$ and $k$ are two constants Substituting $x=x_0e^{kt}$ in eq. $1$, we get $k^2x_0e^{kt}=-\omega^2x_0e^{kt}$ or, $k^2+\omega^2=0$ $(\because e^{kt}\ne0 )$ or, $k=±i\omega$ Thus, the solutions of eq. $1$ are $x_1=x_0^1e^{i\omega t}$ and $x_2=x_0^2e^{-i\omega t}$ Therefore, the general solution of eq. $1$ is given by $x=x_1+x_2$ or, $x=x_0^1e^{i\omega t}+x_0^2e^{-i\omega t}$ or, $x=x_0^1(\cos\omega t+i\sin\omega t)-x_0^2(\cos\omega t+i\sin\omega t)$ or, $x=(x_0^1+x_0^2)\cos\omega t+i(x_0^1-x_0^2)\sin\omega t$ or, $x=B\cos\omega t+A\sin\omega t$ or, $x=A\sin\omega t+B\cos\omega t$ where, $A=i(x_0^1-x_0^2)$ and $B=(x_0^1+x_0^2)$ are two constants.

The equation of motion of a harmonic oscillator is given by $m\frac{d^2x}{dt^2}=–k_fx$ or, $\frac{d^2x}{dt^2}=–\frac{k_f}{m}x$ or, $\frac{d^2x}{dt^2}=–\omega^2x$ $\bigg($where $\omega=\sqrt {\frac{k_f}{m}}$$\bigg)$ ..........$(1)$ Let $x=x_0e^{kt}$ is the solution of eq, $1$, where $A$ and $k$ are two constants Substituting $x=x_0e^{kt}$ in eq. $1$, we get $k^2x_0e^{kt}=-\omega^2x_0e^{kt}$ or, $k^2+\omega^2=0$ $(\because e^{kt}\ne0 )$ or, $k=±i\omega$ Thus, the solutions of eq. $1$ are $x_1=x_0^1e^{i\omega t}$ and $x_2=x_0^2e^{-i\omega t}$ Therefore, the general solution of eq. $1$ is given by $x=x_1+x_2$ or, $x=x_0^1e^{i\omega t}+x_0^2e^{-i\omega t}$ or, $x=x_0^1(\cos\omega t+i\sin\omega t)-x_0^2(\cos\omega t+i\sin\omega t)$ or, $x=(x_0^1+x_0^2)\cos\omega t+i(x_0^1-x_0^2)\sin\omega t$ or, $x=B\cos\omega t+A\sin\omega t$ or, $x=A\sin\omega t+B\cos\omega t$ where, $A=i(x_0^1-x_0^2)$ and $B=(x_0^1+x_0^2)$ are two constants.