A material point of unit mass, m=1 , moves along the x -axis…

A material point of unit mass, $m=1$, moves along the $x$-axis under an elastic (restoring) force $-\omega^2x$ only, with no friction. Find the general law of motion, and the particular motion satisfying $x(0)=1$ (the initial velocity $v_0=\dot x(0)$ left free). $\ddot x + \omega^2x = 0$.

Step 1: Characteristic equation

\[ \lambda^2+\omega^2=0 \;\Longrightarrow\; \lambda=\pm i\omega. \]

Step 2: General solution

\[ x(t)=C_1\cos\omega t+C_2\sin\omega t = A\cos(\omega t-\varphi_0), \qquad A=\sqrt{C_1^2+C_2^2},\ \ \varphi_0=\arctan\frac{C_2}{C_1}. \tag{1} \]

This is simple harmonic motion of period $T=2\pi/\omega$, independent of the initial data.

Step 3: Apply the initial conditions

$x(0)=1$: $C_1=1$. Differentiating (1), $\dot x=-C_1\omega\sin\omega t+C_2\omega\cos\omega t$, so $\dot x(0)=C_2\omega=v_0\Rightarrow C_2=v_0/\omega$.

Step 4: Amplitude and phase

\[ A=\sqrt{1+\frac{v_0^2}{\omega^2}}=\frac{\sqrt{\omega^2+v_0^2}}{\omega}, \qquad \varphi_0=\arctan\frac{v_0}{\omega}. \]

Step 5: Particular solution

Step 6: Physical interpretation

The total mechanical energy $E=\tfrac12\bigl(\dot x^2+\omega^2x^2\bigr)=\tfrac12\omega^2A^2$ is constant along the motion — friction-free elastic motion conserves energy, with the initial velocity $v_0$ trading off against the initial displacement to keep $E$ fixed. The period $T=2\pi/\omega$ depends only on $\omega$ (equivalently, on $m$ and the spring constant), never on the initial conditions $x(0)$, $v_0$.

Step 7: Verification

$\dot x=-\omega\sin\omega t+v_0\cos\omega t$, $\ddot x=-\omega^2\cos\omega t-v_0\omega\sin\omega t=-\omega^2x$, so $\ddot x+\omega^2x=0\ \checkmark$. At $t=0$: $x(0)=1\ \checkmark$, $\dot x(0)=v_0\ \checkmark$.