Simple harmonic motion — phase space, energy diagrams, and examples

Question

What are the key equations and energy relationships in SHM? How do displacement, velocity, acceleration, and energy vary with time and position?

Solution — Step by Step

SHM is defined by:

a = -\omega^2 x

Acceleration is proportional to displacement and directed toward the mean position. The solution is:

x = A\sin(\omega t + \phi)

where $A$ = amplitude, $\omega = 2\pi/T$ = angular frequency, $\phi$ = initial phase.

Velocity: $v = A\omega\cos(\omega t + \phi) = \omega\sqrt{A^2 - x^2}$

Maximum at mean position ( $x = 0$ ): $v_{max} = A\omega$
Zero at extreme positions ( $x = \pm A$ )

Acceleration: $a = -A\omega^2\sin(\omega t + \phi) = -\omega^2 x$

Maximum at extremes ( $x = \pm A$ ): $a_{max} = A\omega^2$
Zero at mean position ( $x = 0$ )

Key insight: velocity and displacement are 90 degrees out of phase. Acceleration and displacement are 180 degrees out of phase (always opposite).

Kinetic energy: $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

Potential energy: $PE = \frac{1}{2}m\omega^2 x^2$

Total energy: $E = KE + PE = \frac{1}{2}m\omega^2 A^2$ (constant)

At the mean position: KE is maximum, PE is zero. At the extremes: PE is maximum, KE is zero. At $x = A/\sqrt{2}$ : KE = PE = $E/2$ .

graph TD
    A[SHM Quantities] --> B[At mean position x=0]
    A --> C[At extreme x=A]
    B --> D["Velocity: maximum = Aw"]
    B --> E["Acceleration: zero"]
    B --> F["KE: maximum, PE: zero"]
    C --> G["Velocity: zero"]
    C --> H["Acceleration: maximum = Aw2"]
    C --> I["KE: zero, PE: maximum"]
    A --> J["At x=A/sqrt2: KE = PE"]

Why This Works

SHM occurs whenever the restoring force is proportional to displacement. Common examples:

System	Angular Frequency $\omega$	Time Period $T$
Spring-mass	$\sqrt{k/m}$	$2\pi\sqrt{m/k}$
Simple pendulum	$\sqrt{g/L}$	$2\pi\sqrt{L/g}$
Physical pendulum	$\sqrt{mgd/I}$	$2\pi\sqrt{I/mgd}$

The energy conservation in SHM is a continuous exchange between KE and PE. The total mechanical energy is constant because there is no dissipation (ideal case). The KE and PE each oscillate at twice the frequency of the displacement — this is because $\sin^2(\omega t)$ and $\cos^2(\omega t)$ have frequency $2\omega$ .

Alternative Method

For JEE problems asking “at what displacement is KE = $n$ times PE?”:

From $KE = n \cdot PE$ :

\frac{1}{2}m\omega^2(A^2 - x^2) = n \cdot \frac{1}{2}m\omega^2 x^2

A^2 - x^2 = nx^2

x = \frac{A}{\sqrt{n+1}}

Memorise: KE = PE at $x = A/\sqrt{2}$ , KE = 3PE at $x = A/2$ .

Common Mistake

Students often write the time period of a simple pendulum as $T = 2\pi\sqrt{g/L}$ (inverting the fraction). The correct formula is $T = 2\pi\sqrt{L/g}$ . Increasing the length increases the time period (longer pendulum swings slower). A quick dimensional check: $\sqrt{L/g}$ has dimensions of time, while $\sqrt{g/L}$ has dimensions of $\text{s}^{-1}$ (frequency). This catches the error immediately.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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