11 - Simple Harmonic Motion

A structured guide to simple harmonic motion, including equilibrium, oscillation, kinematics, energy, mass–spring systems, simple pendulums, and problem-solving methods.

Conditions for

An oscillation is repeated motion back and forth about an . Examples include a cart attached to a spring, a swinging pendulum, and a vibrating string.

The is where the net force is zero. If the equilibrium is stable, displacing the object produces a restoring force that points back toward equilibrium. occurs when the restoring force is proportional to displacement and directed opposite to it:

F_x=-kx.

Using Newton’s second law gives

a_x=-\frac{k}{m}x.

The negative sign is essential: it indicates that force and displacement point in opposite directions. In a general SHM description, the acceleration is written as

a_x=-\omega^2x.

An ideal oscillator has no friction or air resistance, so its remains constant. A real oscillator often undergoes , in which energy is dissipated and the decreases with time.

Takeaway: SHM is identified by a restoring acceleration proportional to displacement and directed toward equilibrium.

Describing the Oscillation Cycle

The $A$ is the greatest displacement from equilibrium. The turning points are $x=+A$ and $x=-A$ . The object stops momentarily at each turning point before reversing direction.

The $T$ is the time for one complete cycle, while frequency $f$ is the number of cycles per unit time:

f=\frac{1}{T}, \qquad T=\frac{1}{f}.

The is

\omega=2\pi f=\frac{2\pi}{T}.

A convenient position model is

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

where the $\phi$ determines the initial point in the cycle. If $\phi=0$ , then $x(0)=A$ , so the object begins at its positive maximum displacement.

Differentiating the position gives

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

and

a(t)=-A\omega^2\cos(\omega t+\phi)=-\omega^2x(t).

Therefore,

v_{\max}=A\omega, \qquad |a|_{\max}=A\omega^2.

The object moves fastest at equilibrium, where $x=0$ , and has zero velocity at the turning points. Acceleration has zero magnitude at equilibrium and maximum magnitude at the turning points.

For an ideal oscillator, motion from $+A$ to equilibrium takes $\frac{T}{4}$ , motion from $+A$ to $-A$ takes $\frac{T}{2}$ , and returning to $+A$ with the same direction of motion takes one full .

Takeaway: Position, velocity, and acceleration describe the same cycle but are shifted in phase.

Energy Exchange in Oscillators

For a horizontal mass–spring oscillator, the spring potential energy is

U_s=\frac{1}{2}kx^2,

and the kinetic energy is

K=\frac{1}{2}mv^2.

The is conserved in the ideal model:

E=K+U_s =\frac{1}{2}mv^2+\frac{1}{2}kx^2.

At maximum displacement, $x=\pm A$ , the speed is zero and all the energy is spring potential energy:

E=\frac{1}{2}kA^2.

At equilibrium, $x=0$ , spring potential energy is zero and kinetic energy is maximum. Thus, the total energy is proportional to the square of . If doubles, total energy becomes four times as large.

Solving the energy equation for speed gives

v=\pm\sqrt{\frac{k}{m}(A^2-x^2)}.

The two signs represent motion through the same position in opposite directions.

For a vertical spring, gravity shifts the . When displacement is measured from the shifted equilibrium, the same SHM equations apply. The is determined by mass and spring constant, not by the amount gravity stretches the spring.

Takeaway: Energy moves between kinetic and potential forms while the ideal total remains constant.

Mass–Spring Oscillators

For a mass $m$ attached to an ideal spring with spring constant $k$ , Newton’s second law gives

-kx=ma_x.

Comparing this with $a_x=-\omega^2x$ produces

\omega=\sqrt{\frac{k}{m}}.

The mass–spring and frequency are therefore

T_s=2\pi\sqrt{\frac{m}{k}},

f_s=\frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Increasing mass increases the according to $T_s\propto\sqrt{m}$ . Increasing the spring constant decreases the according to $T_s\propto\frac{1}{\sqrt{k}}$ . For an ideal spring, the does not depend on .

For example, if $m=0.50\ \text{kg}$ and $k=200\ \text{N/m}$ , then

T=2\pi\sqrt{\frac{0.50}{200}}\approx0.314\ \text{s},

f=\frac{1}{T}\approx3.18\ \text{Hz}.

If $A=0.080\ \text{m}$ , the maximum speed is

v_{\max}=A\omega=A\frac{2\pi}{T}\approx1.60\ \text{m/s}.

Takeaway: The mass and spring constant determine the ; affects maximum speed and energy but not the ideal .

Small-Angle Pendulums

A consists of a point-like bob attached to a massless, inextensible string of length $\ell$ . When displaced by an angle $\theta$ , the tangential component of gravity is the restoring force:

F_t=-mg\sin\theta.

For small angles measured in radians,

\sin\theta\approx\theta.

This makes the restoring effect approximately proportional to angular displacement, so the pendulum approximately undergoes SHM. Its is

T_p=2\pi\sqrt{\frac{\ell}{g}},

and its frequency is

f_p=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}}.

The increases with the square root of length and decreases with the square root of gravitational field strength. It does not depend on bob mass and is approximately independent of only for small oscillations. At larger amplitudes, the actual is somewhat greater than the small-angle prediction.

Pendulum energy alternates between gravitational potential energy and kinetic energy. If the bob rises through height $h$ , then

\Delta U_g=mgh.

For release from rest at angle $\theta_0$ , the height above the lowest point is

h=\ell(1-\cos\theta_0).

At a later angle $\theta$ , conservation of energy gives

mg\ell(1-\cos\theta_0) =\frac{1}{2}mv^2+mg\ell(1-\cos\theta).

The mass cancels, showing that the speed at a given angle is independent of bob mass in the ideal model.

Takeaway: A pendulum behaves as SHM only approximately, and the small-angle condition is central to its standard formula.

Comparing the Two Main Models

Both mass–spring systems and simple pendulums have a stable equilibrium, a restoring effect, maximum speed at equilibrium, and zero speed at maximum displacement. In the ideal model, both conserve total .

For a mass–spring system, the restoring effect is described by and the is

T_s=2\pi\sqrt{\frac{m}{k}}.

Its depends on mass $m$ and spring constant $k$ .

For a small-angle , the restoring effect comes from the tangential component of gravity and the is

T_p=2\pi\sqrt{\frac{\ell}{g}}.

Its depends on length $\ell$ and gravitational field strength $g$ , but not on bob mass.

A useful comparison is:

Spring oscillator: changing $m$ or $k$ changes the .
: changing $\ell$ or $g$ changes the .
Ideal spring: is independent of .
Small-angle pendulum: is approximately independent of .

Takeaway: Identify the restoring mechanism first; it determines which equation applies.

A Reliable Problem-Solving Method

Begin by choosing the and defining displacement from it. Then identify the restoring force and verify that it points toward equilibrium.

Use the appropriate equation:

T_s=2\pi\sqrt{\frac{m}{k}}

for an ideal mass–spring system, or

T_p=2\pi\sqrt{\frac{\ell}{g}}

for a small-angle .

Use the position model and phase when the question concerns initial position or direction of motion:

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

Use energy when the question asks for speed, displacement, or maximum values. For a spring oscillator, begin with

E=\frac{1}{2}mv^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2.

Check the assumptions: the spring should be treated as ideal, damping should be negligible, and the pendulum angle should be small when using the standard pendulum . Finally, check units: is measured in seconds, frequency in hertz, in radians per second, and energy in joules.

Takeaway: A reliable solution identifies the model, defines equilibrium clearly, selects the correct equation, and checks assumptions and units.

Conditions for Simple harmonic motion