17.2 Simple Harmonic Motion

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a special type of periodic (oscillatory) motion in which the acceleration of the body is:

1. Directly proportional to its displacement from the mean (equilibrium) position, and
2. Always directed towards the mean position (opposite to displacement).

Mathematically:

$a\propto -x$

The negative sign indicates that acceleration always acts in the direction opposite to displacement — it is a restoring acceleration.

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Restoring Force in a Mass-Spring System

Consider a mass $m$ attached to a spring of spring constant $k$, resting on a frictionless surface. When displaced by $x$ from equilibrium, Hooke's Law gives the restoring force:

$F=-kx$

Applying Newton's second law ($F=ma$):

$ma=-kx$

$a=-\frac{k}{m}x$

Since $k$ and $m$ are constants, this confirms $a\propto -x$, proving the mass-spring system executes SHM.

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The SHM Equation: $a=-{\omega }^{2}x$

Comparing $a=-\frac{k}{m}x$ with the standard SHM equation:

$\boxed{a=-{\omega }^{2}x}$

we identify:

${\omega }^2=\frac{k}{m}\Rightarrow \omega =\sqrt{\frac{k}{m}}$

where $\omega$ is the angular frequency (rad s⁻¹).

Key values of acceleration in SHM:

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Period of SHM

The time period $T$ is the time for one complete oscillation. Since $\omega =\frac{2\pi }{T}$:

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$

$\boxed{T=2\pi \sqrt{\frac{m}{k}}}$

Key observations:

• $T$ increases if mass $m$ increases (heavier object oscillates more slowly).
• $T$ decreases if spring constant $k$ increases (stiffer spring oscillates faster).
• $T$ is independent of amplitude (isochronous property).

Worked Example

Problem: A mass of $0.5\text{ kg}$ is attached to a spring with $k=200{\text{ N m}}^{-1}$. Find (a) the angular frequency, (b) the period, and (c) the acceleration when $x=0.05\text{ m}$.

Solution:

(a) $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.5}}=\sqrt{400}=20{\text{ rad s}}^{-1}$

(b) $T=\frac{2\pi }{\omega }=\frac{2\pi }{20}\approx 0.314\text{ s}$

(c) $a=-{\omega }^{2}x=-(20{)}^2\times 0.05=-400\times 0.05=-20{\text{ m s}}^{-2}$

The negative sign confirms the acceleration is directed back towards the mean position.

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Summary