17.4 Kinematics of SHM

Kinematics of SHM

Displacement

Equation of Displacement with respect to Time

$x=x_0\sin \omega t$

NOTE: Derived using circular motion

Velocity

Derivation of Velocity with respect to Time

• Velocity $v$ is the rate of change of displacement, i.e. the derivative of displacement with respect to time: $v=\frac{dx}{dt}$
• Since $x=x_0\sin \omega t$: \large\begin{align}v&=\frac{d}{dt}(x_{0}\sin \omega t)\\&=x_{0}\omega \cos \omega t\end{align}
• Recall that linear velocity is given by $v=r\omega$, since in this case the radius is the amplitude, $v_0=x_0\omega$: $v=v_0\cos \omega t$

Derivation of Velocity with respect to Displacement

Let the displacement $x$ and velocity $v$ of a simple harmonic oscillator be:

$$\begin{array}{rl}x& =x_0\sin (\omega t)⟹\sin (\omega t)=\frac{x}{x_0}\\ v& =x_0\omega \cos (\omega t)⟹\cos (\omega t)=\frac{v}{x_0\omega }\end{array}\text{\hspace{1em}(1)(2)}$$

Using the Pythagorean trigonometric identity:

$${\sin }^2(\omega t)+{\cos }^2(\omega t)=1$$

Substitute equations $(1)$ and $(2)$ into the identity:

$${\left(\frac{x}{x_0}\right)}^2+{\left(\frac{v}{x_0\omega }\right)}^2=1$$

Expand the terms:

$$\frac{x^2}{x_0^2}+\frac{v^2}{x_0^2{\omega }^2}=1$$

Multiply the entire equation by $x_0^2{\omega }^2$ to eliminate the denominators:

$$\begin{array}{rl}{\omega }^2{x}^2+v^2& =x_0^2{\omega }^2\\ v^2& =x_0^2{\omega }^2-{\omega }^2{x}^2\\ v^2& ={\omega }^2(x_0^2-x^2)\end{array}$$

Solve for $v$ by taking the square root of both sides:

$$\begin{array}{rl}v& =\pm \sqrt{{\omega }^2(x_0^2-x^2)}\\ v& =\pm \omega \sqrt{x_0^2-x^2}\end{array}$$

Acceleration

Derivation of Acceleration with respect to Time

• Acceleration is the rate of change of velocity, i.e. the derivative of velocity with respect to time: $a=\frac{dv}{dt}$
• Since $v=v_0\cos \omega t$: \large \begin{align}\frac{dv}{dt} &= \frac{d}{dt}(v_{0}\cos \omega t) \\ &= -\omega v_{0}\sin \omega t \end{align}
• Recall that linear velocity is given by $v=r\omega$, since in this case the radius is the amplitude, $v_0=x_0\omega$: \large \begin{align}a&=-\omega \times x_{0}\omega \times \sin \omega t \\ &= -\omega^2x_{0}\sin \omega t\end{align}
• Since $x=x_0\sin \omega t$: $a=-{\omega }^{2}x$