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Topic 16 of 20

Simple Harmonic Motion

Simple harmonic motion (SHM) is the most important type of oscillation in physics. It arises whenever a system experiences a restoring force proportional to displacement from equilibrium — which describes springs, pendulums, sound waves, electromagnetic waves, and many other phenomena. The solution is always sinusoidal.

Key Concepts

Definition of SHM

Motion where the restoring force is proportional to displacement and directed toward equilibrium:

F = -kx

. This produces sinusoidal oscillation:

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

Amplitude and Phase

Amplitude

A

is the maximum displacement from equilibrium. Phase

\phi

sets the initial conditions. Both are determined by initial position and velocity.

Angular Frequency and Period

Spring–mass:

\omega = \sqrt{k/m}

T = 2\pi\sqrt{m/k}

. Period is independent of amplitude — a key property of SHM.

Simple Pendulum

For small angles (

\theta < 15°

T = 2\pi\sqrt{L/g}

. Period depends only on length and

g

, not mass or amplitude. A physical (real) pendulum has

T = 2\pi\sqrt{I/mgd}

Energy in SHM

Total energy

E = \frac{1}{2}kA^2 = \text{const}

. At any displacement

x

K = \frac{1}{2}k(A^2 - x^2)

U = \frac{1}{2}kx^2

. Maximum speed at

x=0

v_\text{max} = A\omega

Resonance

When a driven oscillator is forced at its natural frequency

\omega_0

, the amplitude grows (in the absence of damping, grows without bound). Resonance is responsible for many engineering failures and is exploited in musical instruments.

Key Equations

Equation of motion (SHM)

F = -kx \quad\Longrightarrow\quad \ddot{x} = -\omega^2 x

Linear restoring force; solutions are sines and cosines.

General solution

x(t) = A\cos(\omega t + \phi), \quad v(t) = -A\omega\sin(\omega t + \phi)

A and φ set by initial conditions.

Period (spring–mass)

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

Longer period for heavier mass or softer spring.

Period (simple pendulum)

T = 2\pi\sqrt{\frac{L}{g}}

Valid only for small angles; independent of mass and amplitude.

Energy in SHM

E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv_\text{max}^2 = K + U

Total mechanical energy is conserved; K and U oscillate out of phase.

Worked Example

Spring–Mass on a Frictionless Surface

Problem

A 0.5 kg mass on a spring ( $k = 50$ N/m) is displaced 0.1 m from equilibrium and released from rest. Find the period, maximum speed, and total energy.

Solution

Period:

T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5}{50}} = 2\pi\times 0.1 \approx 0.628 \text{ s}

Total energy ( $A = 0.1$ m):

E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(50)(0.1)^2 = 0.25 \text{ J}

Maximum speed (at $x = 0$ ):

v_\text{max} = A\omega = A\sqrt{\frac{k}{m}} = 0.1\sqrt{\frac{50}{0.5}} = 0.1\times10 = 1.0 \text{ m/s}

Answer T ≈ 0.628 s; E = 0.25 J; v_max = 1.0 m/s.

Practice

Exercises

7 problems

1 of 7

A spring-mass system has $k = 200$ N/m and $m = 0.8$ kg. What is the angular frequency $\omega$ (in rad/s)?

Your answer

rad/s

2 of 7

What is the period (in s) of the spring-mass system in Exercise 1?

Your answer

3 of 7

A mass on a spring oscillates with amplitude $A = 0.15$ m and $\omega = 10$ rad/s. What is the maximum speed (in m/s)?

Your answer

m/s

4 of 7

The same oscillator (amplitude $0.15$ m, $\omega = 10$ rad/s, mass $0.5$ kg) — what is the total mechanical energy (in J)?

Your answer

5 of 7

A simple pendulum has length $L = 1.2$ m. What is its period (in s) for small oscillations? Use $g = 9.8$ m/s².

Your answer

6 of 7

For the oscillator in Exercise 3–4, what is the speed (in m/s) when the displacement is $x = 0.09$ m?

Your answer

m/s

7 of 7

A mass $m = 0.2$ kg hangs from a spring of $k = 80$ N/m and is given an initial displacement of $0.05$ m from rest. What is the maximum kinetic energy (in J) during the oscillation?

Your answer

Key Takeaways

SHM arises whenever $F = -kx$ : the restoring force is proportional to and opposite to displacement.
Period of a spring–mass system depends on $m/k$ , not amplitude — doubling $A$ does not change $T$ .
The simple pendulum period $T = 2\pi\sqrt{L/g}$ is valid only for small angles (less than ~15°).
Energy oscillates between kinetic (max at equilibrium) and potential (max at amplitude); total is constant.
At displacement $x$ : $v = \omega\sqrt{A^2 - x^2}$ — useful for finding speed without using time.