Simple Harmonic Motion

This topic is contained in pages $124-132$ of the textbook.

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A system in SHM is one where an object oscillates about an equilibrium point. The object is acted on by an unbalanced restoring force which acts proportional to the object’s displacement from the rest or equilibrium position and which always acts in the direction of the equilibrium position.

→ This means that when an object is travelling towards the rest position it is accelerating (getting faster) until it reaches its maximum speed at the rest position and when the object is travelling away from the rest position it is decelerating (getting slower) until it comes to a stop, its minimum speed, at its maximum displacement.

→ Furthermore, in objects in SHM have an acceleration proportional to, and in the opposite direction to their displacement from the rest position, i.e. a displacement proportional to and in the opposite direction to the acceleration, as shown by the following equation.

$$ \dfrac{d^2y}{dx^2}=-\omega^2y $$

The displacement of an object in SHM can be expressed by the equations $y=Asin\omega t$ (used when $y=0$ at $t=0$) and $y = Acos \omega t$ (used when $y=A$ at $t=0$).

→ These equations can be used to derive the equation which defines SHM, $a = \omega^2 y$, using calculus.

$v=\dfrac{dy}{dt}$ and $y=Acos\omega t$

$v=\dfrac{d}{dt}(Acos\omega t)$

$v=-A\omega sin\omega t$

$a=\dfrac{dv}{dt}$ and $v=-A\omega sin\omega t$

$a=\dfrac{d}{dt}(-A\omega sin\omega t)$

$a=-A\omega^2 cos\omega t=-\omega^2 (Acos\omega t)$ and $y=Acos\omega t$

$a=\dfrac{d^2y}{dx^2}=-\omega^2 y$

$v=\dfrac{dy}{dt}$ and $y=Asin\omega t$

$v=\dfrac{d}{dt}(Asin\omega t)$

$v=A\omega cos\omega t$

$a=\dfrac{dv}{dt}$ and $v=A\omega cos\omega t$

$a=\dfrac{d}{dt}(A\omega cos\omega t)$

$a=-A\omega^2 sin\omega t=-\omega^2 (Asin\omega t)$ and $y=Asin\omega t$

$a=\dfrac{d^2y}{dx^2}=-\omega^2 y$

→ The object’s acceleration is at its maximum when $y=\pm A$, ($a=\mp \omega A^2$).

Remember $a$ acts to oppose $y$

An alternative expression for $v$ can be derived, using trigonometric substitutions.

$v=-A\omega sin\omega t$

$v=-(\pm A\omega \sqrt{sin^2\omega t})$

$v=-(\pm A\omega \sqrt{1-cos^2\omega t})$ and $y=Acos\omega t$

$v=-(\pm A\omega \sqrt{1-\frac{y^2}{A^2}})$

$v=-(\pm \omega \sqrt{A^2-y^2})$

$v=\mp \omega \sqrt{A^2-y^2}$

$v=A\omega cos\omega t$

$v=\pm A\omega \sqrt{cos^2\omega t}$

$v=\pm A\omega \sqrt{1-sin^2\omega t}$ and $y=Asin\omega t$

$v=\pm A \omega \sqrt{1-\frac{y^2}{A^2}}$

$v=\pm \omega \sqrt{A^2-y^2}$

→ The object’s speed is at its maximum when $y=0$ ⇒ $v_{max}=\pm \omega A$.

The kinetic energy of an object in SHM can be derived using the above expression for $v$.

$E_k=\dfrac{1}{2}mv^2$ and $v=\pm \omega \sqrt{A^2-y^2}$

$E_k=\dfrac{1}{2}m(\omega \sqrt{A^2-y^2})^2$

$E_k=\dfrac{1}{2}m\omega^2 (A^2-y^2)$

→ The object’s kinetic energy is at its maximum when $y=0$ ⇒ $E_{k_{max}}=\dfrac{1}{2}m\omega^2 A^2$