2.2 Beats and Resonance

Attach the mass to a shaker and drive it. With no damping, and with the spring tuned so the system's natural frequency is $2$ , the equation is

$x'' + 4x = \cos(a t),$

where $a$ is the frequency you choose to push at. The method of undetermined coefficients says: guess a response of the same shape as the input, $x_p = A\cos(at)$ . Substituting gives $-a^2 A \cos(at) + 4A\cos(at) = \cos(at)$ , so $A = 1/(4 - a^2)$ — provided $a \ne 2$ . Adding the homogeneous solution and imposing rest initial conditions $x(0) = x'(0) = 0$ produces

$x(t) = \frac{\cos(a t) - \cos(2 t)}{4 - a^2}.$

The denominator is the whole story in miniature. Push at a frequency far from $2$ and the response is small; push near $2$ and $4 - a^2$ is tiny, so the response is enormous. But the shape is stranger than "big" — as the figure shows, the motion swells and collapses in slow packets.

A red wave along the time axis whose amplitude swells and shrinks in slow packets: fast oscillations inside a slowly waving envelope, the classic picture of beats. Dragging the slider a toward the natural frequency 2 stretches the packets longer and taller, so near a = 1.95 a single enormous swell fills the window - the approach to resonance. — The rest-start response of $x'' + 4x = \cos(at)$ , an undamped oscillator with natural frequency $2$ : time $t$ runs along the horizontal axis from $0$ to $40$ and the displacement $x$ is vertical. Dragging the forcing frequency $a$ toward $2$ lengthens and heightens the beat packets — the run-up to resonance.

Explore

Observe. Set $a = 1$ , well away from resonance. The motion is bounded and busy, and it barely leaves the time axis: its swings stay far inside the $\pm 10$ ticks. Keep that scale in mind, and note the coefficient $1/(4 - a^2)$ is only $1/3$ here.
Predict. Move $a$ to $1.6$ . Will the packets get longer, shorter, taller, or shorter and taller? Write your guess down first.
Verify. Drag to $a = 1.6$ , then $a = 1.8$ , then $a = 1.9$ . Record roughly how many packets fit in the window at each setting. The packet count is falling — toward what, as $a \to 2$ ?
Observe. At $a = 1.95$ a single swell nearly fills the window. The amplitude scale has changed dramatically. Compute $1/(4 - a^2)$ at this setting and see whether it matches the height you observe.
Predict, then reason. What does the formula do at exactly $a = 2$ ? The slider stops just short — explain from the algebra why the figure cannot draw the resonant case, and what the true resonant solution must look like instead.

Connect

The product-to-sum identity turns the response into

$x(t) = \frac{2}{4 - a^2}\, \sin\!\left(\frac{2 + a}{2}\,t\right) \sin\!\left(\frac{2 - a}{2}\,t\right),$

and now the two time scales are visible separately. The first sine oscillates at roughly the average frequency — that is the fast wiggle you see. The second oscillates at half the difference $2 - a$ , which is small when $a$ is near $2$ , and it acts as a slowly varying amplitude: the envelope. This is the phenomenon of beats, the same effect a piano tuner hears as a slow throb when a string is close to but not on pitch. As $a \to 2$ the envelope period $4\pi/(2-a)$ grows without bound and the coefficient $2/(4 - a^2)$ grows with it.

At exactly $a = 2$ the guess $A\cos 2t$ fails, because $\cos 2t$ already solves the homogeneous equation — the standard repair is to multiply by $t$ , and the true resonant solution is $x(t) = \tfrac{1}{4} t \sin 2t$ . The envelope is no longer a slow sine but the straight line $\pm t/4$ : amplitude that grows forever. Real systems have damping, which caps the growth at a large but finite value, and real structures are designed to keep their natural frequencies away from the frequencies they will be driven at.

Example 1 How close is too close?

For $x'' + 4x = \cos(at)$ with rest initial conditions, find the forcing frequency $a < 2$ at which the particular-solution coefficient first exceeds $5$ , and find the beat period there.

Show solution

The coefficient is $1/(4 - a^2)$ , and $1/(4 - a^2) > 5$ requires $4 - a^2 < 1/5$ , so $a^2 > 3.8$ and $a > \sqrt{3.8} \approx 1.949$ . The beats (the packets) repeat every $2\pi/(2 - a) = 2\pi/(2 - \sqrt{3.8}) \approx 124$ , and the envelope sine itself has period $4\pi/(2 - a) \approx 248$ — both vastly longer than the window in the figure, which is why at $a = 1.95$ you see a single rising swell rather than a repeating pattern. Detuning by less than $3\%$ has both pushed the coefficient past $5$ — twenty times the static deflection $1/4$ that the same unit force produces at $a = 0$ — and pushed the beat period past anything an observer would recognize as a beat.

An original work of XYZ Homework, built around interactive XYZ 3D figures, following the standard first-course sequence (first-order equations through systems and the phase plane). Distinct from the LibreTexts-sourced Differential Equations for Engineers (Lebl) edition, which has its own attribution. License: CC-BY-NC-SA-4.0.