A mass on a spring oscillates sinusoidally without going round anything. Covers Hooke's law and the equilibrium position from which all displacement is measured, the equation for free undamped harmonic motion with omega determined by the apparatus and the amplitude and phase by the initial conditions, and then the three ways the idealisation is relaxed: damped motion with a decaying envelope, resonance where forcing at the natural frequency makes the envelope grow, and beats where forcing at a different frequency produces a slow modulation that sum-to-product exposes.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.1 Applications of Sinusoids, pp. 885-891
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 885-891 — the pages these objectives are drawn from
Warm-up
Every sinusoid so far described something rotating. A bouncing spring is not rotating, and the previous lesson's machinery applies to it anyway.
Discussion prompt
A mass bounces on a spring, going 5 centimetres below its rest position and 5 centimetres above, taking 2 seconds for a full bounce. Without any physics, name the amplitude, the baseline and the period.
Hint: The three questions are the same ones as for a Ferris wheel.
Answer:
Amplitude 5 centimetres, baseline zero since the displacement is measured from the rest position, and period 2 seconds, giving omega equal to pi.
\[ x(t) = 5\sin(\pi t + \phi) \]
So the modelling is identical to last lesson's. What is new is that the period is not something you measure and put in — it is determined by the spring and the mass, and can be predicted before the thing is ever set moving.
Concept
For a mass on a spring the angular frequency is fixed by the spring constant and the mass alone. The amplitude and phase are fixed by how the motion was started, and have no effect on the frequency at all.
\[ x(t) = A\sin(\omega t + \phi), \qquad \omega = \sqrt{\frac{k}{m}} \]
That separation is the striking fact about oscillation: pull the mass down further and it swings further but takes exactly as long. The same is true of a pendulum for small swings, and it is why pendulum clocks work.
Figure (svg): The equation for free undamped harmonic motion, with the formulas for omega from the spring constant and mass, for the amplitude from the initial displacement and velocity, and the two conditions determining the phase
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 887-887
Section
Section 1
Concept
Hanging a mass on a spring stretches it until the spring's restoring force balances the weight. That position is called equilibrium, and every displacement in this model is measured from it rather than from the spring's natural length.
Hooke's law — The force a spring exerts is proportional to how far it is stretched: F equals k times d, where k is the spring constant. A stiffer spring has a larger k.
The sign convention is the opposite of the usual one for a graph and is worth writing down explicitly at the start of any problem. A negative velocity means moving upward.
Figure (svg): A mass hanging on a spring shown in three states: the spring unstretched, the mass at equilibrium where the spring force balances the weight, and the mass displaced below equilibrium with the displacement x marked
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 886-887
Picture it
Down is positive. Getting this backwards flips the phase and produces a motion that is a mirror image of the intended one.
Figure (svg): Two columns giving the sign convention for the initial displacement and the initial velocity in this model
Writing the four sign meanings at the top of the page costs one line and prevents an error that is otherwise invisible until the answer is checked physically.
Worked example
Example 11.1.3's setup. A 64 pound object stretches the spring 8 feet.
\[ \text{An object weighing } 64 \text{ lb stretches a spring } 8 \text{ ft. Find } k \text{ and } m. \]
Apply Hooke's law at equilibrium
Why: The weight equals the spring force there.
\[ 64 = k(8) \]
Solve for the spring constant
Why: Divide by the stretch.
\[ k = 8 \text{lb}\text{ per } \text{ft} \]
Find the mass from the weight
Why: Weight equals mass times g, and g is 32 in these units.
\[ 64 = m(32) \]
Solve for the mass
Why: Divide by 32.
\[ m = 2\text{ slugs} \]
Figure (svg): The solution to Worked example find the spring constant shown as a ladder of expressions, one row per legal move
\[ k = \frac{64}{8} = 8 \text{ lb/ft}, \qquad m = \frac{64}{32} = 2 \text{ slugs} \]
Verify: compute omega and sanity-check it
Why: Omega is the square root of k over m, which is the root of 8 over 2, namely 2 per second. The period is then two pi over 2, about 3.14 seconds — a plausible bounce for a heavy mass on a soft spring.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 887-888
Sorting
The formula needs mass; problems often state weight.
Sort into buckets
Sort each quantity.
Worked example
The structure is identical; only the value of g changes.
\[ \text{A } 2 \text{ kg mass stretches a spring } 0.4 \text{ m. Find } k \text{ and } \omega. \]
Find the weight
Why: Mass times g, with g equal to 9.8 in SI.
\[ w = 2(9.8) = 19.6 N \]
Apply Hooke's law
Why: The weight equals k times the stretch.
\[ 19.6 = k(0.4) \]
Solve for the spring constant
Why: Divide by 0.4.
\[ k = 49 N\text{ per } m \]
Compute omega
Why: The square root of k over m.
\[ \omega = \sqrt{\frac{49}{2}} = 4.95 \]
Figure (svg): The solution to Worked example the same problem in SI units shown as a ladder of expressions, one row per legal move
\[ k = 49 \text{ N/m}, \qquad \omega = \sqrt{\frac{49}{2}} \approx 4.95 \text{ s}^{-1} \]
Verify: check the period
Why: Two pi over 4.95 is about 1.27 seconds, so the mass bounces a little under once a second. That is a reasonable rate for a 2 kg mass hanging on a spring that it stretches by 40 centimetres.
Trap
\[ \text{a } 64 \text{ lb object} \;\Longrightarrow\; m = 64 \;\Longrightarrow\; \omega = \sqrt{\tfrac{8}{64}} = \tfrac{1}{2\sqrt{2}} \]
Take the stated number as the mass
Why: Pounds are the unit people quote, and in ordinary speech weight and mass are used interchangeably.
But pounds measure force, not mass. The mass is the weight divided by g, which is 64 over 32, namely 2 slugs — and omega comes out as 2 rather than about 0.35.
\[ m = \frac{w}{g} = \frac{64}{32} = 2 \text{ slugs} \;\Longrightarrow\; \omega = \sqrt{\tfrac{8}{2}} = 2 \]
Convert weight to mass before using it
Why: The formula for omega needs mass, and a weight in pounds or newtons is not one.
SI users are less exposed to this because kilograms genuinely are mass, but the same trap appears in reverse: a weight in newtons must be divided by 9.8 before it can be used as a mass. The check is dimensional — omega must come out with units of one over time, and it only does when m is a mass.
Fill the middle
A 96 pound object stretches a spring 6 feet. Find omega.
Fill in the blanks
k = \frac163 = 4/sqrt3, \quad m = \frac______ = ___ \;\Longrightarrow\; \omega = \sqrt___}}___}} = ___
Why: Hooke's law gives k as 16 pounds per foot and the weight-to-mass conversion gives 3 slugs. Omega is the square root of sixteen thirds, which is 4 over root three, about 2.31 per second — a period of about 2.7 seconds. Both conversions were needed, and skipping either changes omega substantially.
Prediction
Two identical springs hold masses of 2 kg and 8 kg respectively.
Predict first
How do their periods compare?
Correct: The heavier one has twice the period.
Why: Omega is the square root of k over m, so quadrupling the mass halves omega and therefore doubles the period. The square root is doing the work: the period is proportional to the square root of the mass, not to the mass itself. This is why a heavier mass bounces more slowly but not proportionally more slowly, and it is worth knowing as a check on any numerical answer.
Socratic
The model measures displacement from equilibrium rather than from the spring's natural length.
Discussion prompt
What would go wrong if displacement were measured from the natural length instead?
Hint: What would the baseline of the sinusoid be in each case?
Answer:
Measuring from the natural length would make the oscillation happen about a nonzero baseline — the equilibrium stretch — so the model would need an extra constant term B, and every formula would carry it.
Worse, the weight would still be acting, so the restoring force would not be simply proportional to the displacement and the motion would not obviously be sinusoidal at all. Measuring from equilibrium is what makes the weight cancel out of the problem entirely, leaving a pure restoring force.
So it is not a convenience but a simplification with real content: the choice of origin is what makes the equation simple, and choosing it well is part of setting up any physical model.
Section
Section 2
Concept
The amplitude comes from a formula combining both initial conditions. The phase needs both as well, because either one alone leaves two candidate angles and only their combination picks the right one.
Theorem 11.1 — For a mass m on a spring of constant k with initial displacement x0 and initial velocity v0, the displacement at time t is A sine of omega t plus phi, with omega the square root of k over m and A and phi given by the formulas above.
\[ A = \sqrt{x_0^2 + \left(\frac{v_0}{\omega}\right)^2}, \qquad A\sin\phi = x_0, \quad A\omega\cos\phi = v_0 \]
This is the same structure as fitting a sinusoid to two facts in the previous lesson: one equation determines the size and the pair together determine the position in the cycle.
Figure (svg): The equation for free undamped harmonic motion, with the formulas for omega from the spring constant and mass, for the amplitude from the initial displacement and velocity, and the two conditions determining the phase
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 887-887
Picture it
Three formulas, and it is worth noticing which inputs each one uses.
Figure (svg): The equation for free undamped harmonic motion, with the formulas for omega from the spring constant and mass, for the amplitude from the initial displacement and velocity, and the two conditions determining the phase
Only omega uses k and m. Only A and phi use x0 and v0. The two halves of the problem do not interact, which is what makes the frequency independent of the amplitude.
Worked example
Example 11.1.3, part 1. Released 3 feet below equilibrium with no initial velocity.
\[ \text{With } k = 8, \; m = 2, \; x_0 = 3, \; v_0 = 0, \text{ find } x(t) \text{ and the first equilibrium crossing.} \]
Compute omega
Why: The square root of k over m.
\[ \omega = \sqrt{\frac{8}{2}} = 2 \]
Compute the amplitude
Why: With v0 zero the formula reduces to the initial displacement.
\[ A = 3 \]
Find the phase from both conditions
Why: Three sine phi equals 3 gives sine phi equal to 1; six cosine phi equals 0 agrees.
\[ \phi = \frac{\pi}{2} \]
Solve for the first equilibrium crossing
Why: Set x equal to zero and take the smallest positive t.
\[ t = \frac{\pi}{4} \]
Figure (svg): The solution to Worked example released from rest shown as a ladder of expressions, one row per legal move
\[ x(t) = 3\sin\left(2t + \frac{\pi}{2}\right), \qquad t = \frac{\pi}{4} \approx 0.78 \text{ s} \]
Verify: check the direction of travel
Why: Just after pi over four the displacement becomes negative, meaning the object is above equilibrium — so it is travelling upward as it passes through. That is what common sense predicts for something released below equilibrium, and it confirms the sign convention was applied correctly.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 888-888
Prediction
An object is released from equilibrium with a downward velocity.
Predict first
What is its initial displacement, and what does that make the phase?
Correct: x0 = 0 and phi = 0.
Why: Released from equilibrium means the initial displacement is zero, so A sine phi is zero and the phase is zero or pi. The velocity condition decides: a downward velocity is positive, so A omega cosine phi is positive, and the cosine is positive at zero rather than at pi. The phase is zero, which makes the motion a pure sine with no shift — the simplest possible case, and the one that gives the sine function its name in this context.
Worked example
Example 11.1.3, part 2. Same release point, but moving upward at 8 feet per second.
\[ \text{With } x_0 = 3 \text{ and } v_0 = -8, \text{ find } x(t). \]
Note omega is unchanged
Why: The apparatus has not changed.
\[ \omega = 2 \]
Compute the amplitude
Why: Three squared plus negative four squared, under a root.
\[ A = \sqrt{9 + 16} = 5 \]
Use the first condition
Why: Five sine phi equals 3.
\[ \sin \phi = \frac{3}{5} \]
Use the second to pick the quadrant
Why: Ten cosine phi equals negative 8, so the cosine is negative.
Figure (svg): The solution to Worked example released with a velocity shown as a ladder of expressions, one row per legal move
\[ x(t) = 5\sin\big(2t + \phi\big), \quad \sin\phi = \tfrac{3}{5}, \; \cos\phi = -\tfrac{4}{5} \]
Verify: check the amplitude grew
Why: Giving the object an initial velocity made the amplitude grow from 3 to 5, which is right: it starts 3 feet from equilibrium and is already moving, so it will travel further than 3 feet before turning around. An initial velocity can only increase the amplitude, never decrease it, since it enters the formula as a square.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 888-888
Error analysis
A student finds the phase from the displacement condition alone.
Annotate
On: \( 5\sin\phi = 3 \;\Longrightarrow\; \sin\phi = \tfrac{3}{5} \;\Longrightarrow\; \phi = \arcsin\left(\tfrac{3}{5}\right) \approx 0.6435 \)
Both conditions are needed, and they are needed for exactly the reason the arcsine's range exists: one equation in a sine leaves two candidates, and the cosine condition is what chooses. The same structure appeared throughout Lesson 10.2b.
Faded example
An object with omega equal to 3 is released 4 units below equilibrium moving downward at 9 units per second.
Fill in the blanks
A = \sqrt95\right)^2} = \sqrt___}} = ___
Why: The velocity contributes v0 over omega, which is 9 over 3, namely 3, and squaring gives 9. Sixteen plus nine is twenty-five, whose root is 5. Note that the amplitude of 5 exceeds the initial displacement of 4, as it must whenever the object is already moving when released.
Sorting
The apparatus and the release affect different things.
Sort into buckets
Sort each input by what it determines.
Counterexample
A student proposes: the amplitude of the motion equals the distance the object was pulled from equilibrium.
Discussion prompt
Give a case where this fails and explain what the amplitude actually depends on.
Hint: What if the object is not simply released?
Answer:
\[ x_0 = 3, \; v_0 = -8, \; \omega = 2 \;\Longrightarrow\; A = \sqrt{9 + 16} = 5 \ne 3 \]
If the object is given an initial push rather than simply released, it arrives at its turning point further out than where it started. Here it was pulled 3 feet down and pushed upward, and it travels 5 feet from equilibrium in each direction.
The claim is true only in the special case v0 equal to zero, where the formula collapses to A equal to the absolute value of x0. In general the amplitude combines both the initial position and the initial energy of motion, which is why the velocity enters as a square alongside the displacement.
Section
Section 3
Concept
Real oscillations lose energy to friction and air resistance, and their swings shrink. That is modelled by replacing the constant amplitude with a decaying function of time, usually an exponential.
\[ x(t) = A(t)\sin(\omega t + \phi), \qquad A(t) = A_0 e^{-ct} \]
The technique for putting such an expression into the standard form is unchanged: factor out whatever multiplies both terms, collapse the remaining cosine plus sine into a single sinusoid, and the factored part becomes the amplitude function.
Figure (svg): A damped oscillation, with the sinusoid drawn inside a decaying exponential envelope that shrinks towards zero
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 889-889
Picture it
Free, damped and resonant motion differ only in what the amplitude function does.
Figure (svg): The three kinds of oscillation compared: free undamped with constant amplitude, damped with a shrinking envelope, and resonant with a growing one
Writing the general form as A of t times a sine is what unifies them. A constant A is the idealisation; a decaying one is friction; a growing one is resonance.
Worked example
Example 11.1.4, part 1. Factor first, then use the sinusoid method.
\[ \text{Write } x(t) = 5e^{-t/5}\cos(t) + 5e^{-t/5}\sqrt{3}\sin(t) \text{ as } A(t)\sin(\omega t + \phi). \]
Factor out what is common
Why: Both terms carry five e to the minus t over five.
\[ 5 e ^{-\frac{t}{5}} [\cos t + \sqrt{3} \sin t] \]
Collapse the bracket into one sinusoid
Why: Amplitude is the root of one plus three, which is 2.
\[ = 2 \sin(t + \frac{\pi}{3}) \]
Recombine
Why: The factored part multiplies the sinusoid.
\[ x(t) = 10 e ^{-\frac{t}{5}} \sin(t + \frac{\pi}{3}) \]
Identify the amplitude function
Why: Everything multiplying the sine.
\[ A(t) = 10 e ^{-\frac{t}{5}} \]
Figure (svg): The solution to Worked example collapse a damped expression shown as a ladder of expressions, one row per legal move
\[ x(t) = 10e^{-t/5}\sin\left(t + \frac{\pi}{3}\right) \]
Verify: describe the behaviour
Why: As t grows the exponential shrinks to zero, so the amplitude does too and the oscillation dies away. The frequency is unchanged throughout — the wave crosses zero at the same regular intervals however small it becomes, which is exactly what the picture shows.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 889-889
Prediction
A damped oscillation has amplitude function 8 times e to the minus t over 3.
Predict first
What is its amplitude at t equal to zero, and what does it approach?
Correct: 8, approaching zero.
Why: At t equal to zero the exponential is 1, so the amplitude is 8. As t grows the exponent becomes large and negative, so the exponential approaches zero and the amplitude does too. The motion never quite stops in this model, though it becomes unmeasurably small — which is a known limitation of the idealisation rather than a physical claim.
Worked example
A natural question about a damped oscillation, and an ordinary exponential equation.
\[ \text{For } A(t) = 10e^{-t/5}, \text{ find when the amplitude has fallen to } 5. \]
Set up the equation
Why: The amplitude function equals half its initial value.
\[ 10 e ^{-\frac{t}{5}} = 5 \]
Isolate the exponential
Why: Divide by 10.
\[ e ^{-\frac{t}{5}} = \frac{1}{2} \]
Take logarithms
Why: The natural logarithm undoes the exponential.
\[ -\frac{t}{5} = \ln(\frac{1}{2}) \]
Solve
Why: Multiply by negative five and simplify the logarithm.
\[ t = 5 \ln 2 = 3.47 \]
Figure (svg): The solution to Worked example find when the amplitude has halved shown as a ladder of expressions, one row per legal move
\[ t = 5\ln(2) \approx 3.47 \]
Verify: check the half-life is constant
Why: The same calculation starting from any time gives the same interval, since the exponential's decay rate does not change. So the amplitude halves every 3.47 seconds indefinitely — which is the defining property of exponential decay and is what makes the envelope's shape so distinctive.
Trap
A student reasons: as the swings get smaller they must also get quicker, since the object has less distance to cover.
But the graph shows the zero crossings evenly spaced throughout, at the same interval whether the amplitude is 10 or 0.1.
\[ x(t) = A(t)\sin(\omega t + \phi) \quad \text{with } \omega \text{ constant} \]
Read the model: only A depends on t
Why: The argument of the sine is unchanged, so the timing of every crossing and every turning point is unchanged.
Physically the reasoning fails because a smaller swing also means a weaker restoring force, so the object moves more slowly in exact proportion and takes the same time. This is the same fact as the amplitude not affecting the period in the undamped case, and it is why a pendulum clock keeps time even as its swing decays.
Faded example
Write 3 e to the minus t times cosine 2t, plus 3 e to the minus t times sine 2t, in standard form.
Fill in the blanks
3e^sqrt2\big[\cos 2t + \sin 2t\big] = 3e^pi/4 \cdot ___\sin\left(2t + ___\right)
Why: The bracket is a cosine plus a sine with equal coefficients, so its amplitude is the root of one plus one, namely root two, and its phase is pi over four. The full amplitude function is therefore three root two times e to the minus t, and the frequency remains 2 throughout.
Matching
The amplitude function's shape names the kind of motion.
Match the pairs
Why: A constant amplitude is the idealisation with no energy loss or gain. A decaying exponential is friction removing energy. A growing function is energy being fed in faster than it is lost. And a slowly oscillating amplitude is the beat pattern, where two frequencies drift in and out of step — all four are the same equation with a different A of t.
Real world
A car's suspension is designed to damp the oscillation after the car goes over a bump. A worn shock absorber damps less.
Discussion prompt
Describe the difference in the two amplitude functions, and explain why an undamped suspension would be dangerous rather than merely uncomfortable.
Hint: How many bounces does each produce, and what is the car doing between them?
Answer:
A healthy shock absorber gives a rapidly decaying amplitude function, so the car settles within about one bounce. A worn one decays slowly, so the car continues bouncing for several cycles.
The danger is not the discomfort. At the top of each bounce the tyre is carrying less load and gripping less, and in the worst case leaves the road entirely. A car that is still oscillating when it reaches the next bump — or the next corner — has intermittent grip, and braking distances lengthen sharply.
So the design goal is a specific damping rate: enough that the amplitude function falls essentially to zero within one cycle, and not so much that the suspension cannot move at all. The engineering problem is choosing the decay constant, and the model in this lesson is what that choice is made against.
Section
Section 4
Concept
If the whole apparatus is itself oscillated at the natural frequency of the mass and spring, energy is fed in on every cycle and the amplitude grows without bound. That is resonance.
Resonance requires the driving frequency to match the natural frequency. Driving at any other frequency produces beats instead, which is the subject of the next section.
Figure (svg): A forced oscillation at the natural frequency, with the sinusoid drawn inside a linearly growing envelope
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 889-890
Picture it
The mirror image of damping, and produced by the same algebra.
Figure (svg): A forced oscillation at the natural frequency, with the sinusoid drawn inside a linearly growing envelope
Notice that the wave still crosses zero at the same regular intervals. Only the height of each successive swing has changed.
Worked example
Example 11.1.4, part 2. Identical method to the damped case.
\[ \text{Write } x(t) = (t+3)\sqrt{2}\cos(2t) + (t+3)\sqrt{2}\sin(2t) \text{ in standard form.} \]
Factor out the common part
Why: Both terms carry the quantity t plus three, times root two.
\[ (t + 3) \sqrt{2} [\cos 2 t + \sin 2 t] \]
Collapse the bracket
Why: Equal coefficients give amplitude root two and phase pi over four.
\[ = \sqrt{2} \sin(2 t + \frac{\pi}{4}) \]
Recombine
Why: Root two times root two is 2.
\[ x(t) = 2(t + 3) \sin(2 t + \frac{\pi}{4}) \]
Identify the amplitude function
Why: It grows linearly without bound.
\[ A(t) = 2(t + 3) \]
Figure (svg): The solution to Worked example collapse a resonant expression shown as a ladder of expressions, one row per legal move
\[ x(t) = 2(t + 3)\sin\left(2t + \frac{\pi}{4}\right) \]
Verify: evaluate the amplitude at two times
Why: At t equal to zero the amplitude is 6; at t equal to 10 it is 26. The swings are more than four times larger after ten seconds and continue growing, which is the unbounded behaviour the model predicts and which no real spring would survive.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 890-890
Prediction
A structure has a natural frequency of 2 hertz and is driven by an external oscillation.
Predict first
At what driving frequency is resonance expected?
Correct: 2 hertz.
Why: Resonance occurs when the driving frequency matches the system's natural frequency, so that energy is added in step with the motion on every cycle. Driving at 1 or 4 hertz feeds energy in sometimes with and sometimes against the motion, so the effects largely cancel and produce beats rather than growth. This is why bridges are tested against the frequencies of marching or of expected wind loading in particular.
Worked example
The practical question about a resonant system: when does it break?
\[ \text{For } A(t) = 2(t+3), \text{ find when the swing first exceeds } 40 \text{ units.} \]
Set up the inequality
Why: The amplitude function exceeds 40.
\[ 2(t + 3) > 40 \]
Divide by 2
Why: Isolating the bracket.
\[ t + 3 > 20 \]
Solve
Why: Subtract three.
\[ t > 17 \]
Interpret
Why: After 17 seconds the swings exceed the stated limit and keep growing.
\[ t > 17\text{ seconds} \]
Figure (svg): The solution to Worked example when does the amplitude reach a limit shown as a ladder of expressions, one row per legal move
\[ 2(t + 3) > 40 \;\Longrightarrow\; t > 17 \]
Verify: contrast with the damped case
Why: In a damped system the amplitude would approach zero and never exceed any bound after some point. Here it exceeds every bound eventually, so the question always has an answer — which is precisely why resonance is dangerous and why designers avoid driving a structure at its natural frequency.
Error analysis
A student describes a resonant graph.
Annotate
On: \( x(t) = 2(t+3)\sin\left(2t + \tfrac{\pi}{4}\right) \;\longrightarrow\; \text{the frequency increases with } t \)
A growing amplitude makes a wave look faster because its slope increases, but slope and frequency are different things. Checking the spacing of the zero crossings is the direct test and it takes one measurement.
Discrimination
The amplitude function tells you which case you are in.
Sort into buckets
Sort each amplitude function.
Faded example
Write t cosine 3t plus t root three sine 3t in standard form.
Fill in the blanks
t\big[\cos 3t + \sqrt2\sin 3t\big] = t \cdot pi/6\sin\left(3t + ___\right)
Why: The bracket has coefficients 1 and root three, so its amplitude is the root of one plus three, namely 2. The phase satisfies sine phi equal to one half and cosine phi equal to root three over two, giving pi over six. The amplitude function is 2t, growing linearly — a resonant system starting from rest at the origin.
Edge cases
The resonant model predicts an amplitude growing without bound.
Discussion prompt
Is that physically possible? What does the unbounded prediction actually tell an engineer?
Hint: What assumptions does the model make that stop being true for large swings?
Answer:
It is not physically possible. Two assumptions fail: Hooke's law is only linear for modest stretches, and every real system has some damping that grows with speed. Long before the amplitude became infinite, the spring would deform permanently or the system would break.
What the prediction tells an engineer is that there is no equilibrium to settle into. In a damped system the amplitude approaches a finite steady value however hard it is driven; at resonance in an undamped model there is no such value, so the response is limited only by whatever nonlinearity or failure intervenes first.
That is a genuinely useful conclusion even though the number is wrong: it says the design must avoid the resonant frequency rather than merely accommodate it, because no amount of strengthening produces a safe steady state.
Section
Section 5
Concept
If a system is driven at a frequency different from its natural one, the two oscillations drift in and out of step. The result is not resonance but beats: swings that wax and wane periodically.
This is the same computation as the two-tuning-fork example in Lesson 10.4c, arrived at from a physical direction rather than an algebraic one.
Figure (svg): Two sinusoids of nearby frequencies subtracted, producing a rapid oscillation inside a slowly varying envelope, with the sum-to-product form written underneath
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 890-890
Picture it
A fast oscillation inside a slow envelope, and the two frequencies are close together.
Figure (svg): Two sinusoids of nearby frequencies subtracted, producing a rapid oscillation inside a slowly varying envelope, with the sum-to-product form written underneath
The dashed envelope is the slow factor from the sum-to-product form. It is not a separate oscillation, just the amplitude of the fast one varying.
Worked example
Example 11.1.4, part 3. The ratio of the frequencies decides it.
\[ \text{Find the period of } x(t) = 5\sin(6t) - 5\sin(8t). \]
Find each term's period
Why: Two pi over the coefficient.
\[ \frac{\pi}{3}\text{ and } \frac{\pi}{4} \]
Take the ratio of the frequencies in lowest terms
Why: Six to eight reduces to three to four.
\[ 3: 4 \]
Interpret the ratio
Why: For every 3 cycles of the first, the second makes 4.
\[ 3\text{ cycles of the first} \]
Compute the combined period
Why: Three times the first term's period.
\[ 3 \times \pi / 3 = \pi \]
Figure (svg): The solution to Worked example the period of a sum shown as a ladder of expressions, one row per legal move
\[ \text{period} = 3 \cdot \frac{\pi}{3} = 4 \cdot \frac{\pi}{4} = \pi \]
Verify: check both terms close together
Why: After an interval of pi the first term has completed 3 whole cycles and the second 4, so both return to their starting values simultaneously — and pi is the shortest such interval, since no smaller multiple of both periods exists. That is exactly the definition of the combined period.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 890-890
Fill the middle
Find the period of sin(4t) plus sin(6t).
Fill in the blanks
\text2 \fracpi___ = \frac______ \;\Longrightarrow\; \text___ = ___ \cdot \frac______ = ___
Why: The frequencies are in the ratio two to three, so the first term makes 2 cycles while the second makes 3. Two of the first term's periods is two times pi over two, which is pi. Checking against the second term: pi divided by two pi over six is 3, a whole number. Both return together after pi.
Worked example
Sum-to-product turns the sum into a product and reveals the beat structure.
\[ \text{Rewrite } 5\sin(6t) - 5\sin(8t) \text{ as a product and identify the envelope.} \]
Factor out the 5
Why: Both terms share it.
\[ 5 [\sin 6 t - \sin 8 t] \]
Apply the sine difference sum-to-product formula
Why: Half-difference negative t, half-sum 7t.
\[ 5 [2 \sin(-t) \cos(7 t)] \]
Clean up the negative angle
Why: Sine is odd, so the minus comes out.
\[ = -10 \sin(t) \cos(7 t) \]
Identify the two factors
Why: A slow sine and a fast cosine.
\[ \text{envelope } -10 \sin t \]
Figure (svg): The solution to Worked example expose the envelope shown as a ladder of expressions, one row per legal move
\[ x(t) = -10\sin(t)\cos(7t) \]
Verify: check the envelope's period against the answer
Why: The slow factor has period two pi, but the envelope's magnitude peaks twice per cycle, so the beat pattern repeats every pi — matching the period computed the other way. The two routes agree, and the product form is what makes the envelope visible.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 890-890
Trap
\[ \text{periods } \tfrac{\pi}{3} \text{ and } \tfrac{\pi}{4} \;\Longrightarrow\; \text{combined period } = \tfrac{\pi}{3} + \tfrac{\pi}{4} = \tfrac{7\pi}{12} \]
Add the two periods
Why: Two things are being combined, so adding their periods looks like the natural operation.
But after seven pi over twelve, neither term has completed a whole number of cycles, so neither has returned to its starting value. The sum has not repeated.
\[ \text{ratio } \tfrac{6}{8} = \tfrac{3}{4} \;\Longrightarrow\; \text{period} = 3 \cdot \tfrac{\pi}{3} = \pi \]
Find the shortest interval containing a whole number of BOTH cycles
Why: That is a least common multiple, not a sum.
The right mental model is a least common multiple of the two periods, which the frequency ratio computes directly. Checking is easy: divide the candidate period by each individual period and confirm both quotients are whole numbers. Pi over one third is 3 and pi over one quarter is 4 — both whole, so pi works.
Prediction
Two tuning forks at 300 and 304 hertz are struck together.
Predict first
What frequency does the throbbing occur at?
Correct: 4 hertz.
Why: Sum-to-product gives an envelope oscillating at half the frequency difference, namely 2 hertz, but the envelope's magnitude peaks twice per cycle, so the audible throbbing occurs at 4 hertz — the full difference of the two frequencies. The perceived tone is at the average, 302 hertz. This is precisely how a piano tuner works: count the beats per second and it equals the frequency error.
Sorting
Matching or mismatched frequencies give completely different behaviour.
Sort into buckets
Sort each scenario.
Explain it
A classmate cannot see why driving a swing at its own rhythm works so much better than driving it slightly faster.
Discussion prompt
In four sentences or fewer, explain the difference using the two models from this lesson.
Hint: What happens to the pushes over many cycles in each case?
Answer:
A usable answer: if you push in time with the swing, every push adds energy, because the swing is always moving away from you when you push. The amplitude grows on every cycle, which is resonance, and that is why a child on a swing can be got very high with small pushes.
If you push slightly out of time, you are sometimes pushing with the motion and sometimes against it. Over many cycles those cancel out, and the swing gets higher for a while and then lower again — the beats pattern. Matching the frequency is what makes the pushes add up instead of averaging out.
Comparison
Fill the blanks from memory. One equation, four amplitude behaviours.
Comparison matrix
| Kind | Amplitude function | Physical cause |
|---|---|---|
| Free undamped | constant | no energy enters or leaves |
| Damped | decaying, usually exponentially | friction removes energy |
| Resonant | growing without bound | driven at the natural frequency |
| Beats | slowly oscillating | driven at a nearby frequency |
In every one of the four the frequency of the underlying oscillation is unchanged. Everything that distinguishes them lives in the amplitude.
Pattern
Whether the problem asks for an equation of motion or for a description of one, the same five moves cover it.
For a sum of two sinusoids of different frequencies, the period is a least common multiple rather than a sum, found from the ratio of the frequencies in lowest terms.
OpenStax Algebra and Trigonometry 2e, §8.1 Graphs of the Sine and Cosine Functions §8.1
Check
The apparatus determines the frequency.
Check your understanding
A 4 kg mass hangs on a spring with constant 36 newtons per metre. What is omega?
Answer: B
Why: Omega is the square root of k over m, which is the square root of 36 over 4, namely the square root of 9, which is 3 per second. The period is then two pi over 3, about 2.09 seconds.
Check
Amplitude from both initial conditions.
Check your understanding
An object with omega equal to 5 is released 12 units below equilibrium with a downward velocity of 45 units per second. What is the amplitude?
Answer: B
Why: The velocity contributes v0 over omega, which is 45 over 5, namely 9. The amplitude is the square root of 12 squared plus 9 squared, which is the root of 144 plus 81, the root of 225, namely 15. Nine, twelve and fifteen form a Pythagorean triple.
Check
The period of a sum.
Check your understanding
What is the period of sin(2t) plus sin(3t)?
Answer: B
Why: The two periods are pi and two pi over three. The frequencies are in the ratio two to three, so the first term makes 2 cycles while the second makes 3, and two of the first term's periods is two pi. Checking, two pi divided by two pi over three is 3, a whole number, so both terms return together.
Real world
The Millennium Bridge in London was closed two days after opening in 2000 because it swayed alarmingly when crowded. Pedestrians instinctively adjusted their steps to match the sway, and the bridge had a lateral natural frequency close to 1 hertz — near the rate at which people walk.
Discussion prompt
Explain what happened in terms of this lesson's three cases, and say why the eventual fix was to add dampers rather than to stiffen the bridge.
Hint: What did the pedestrians' adjustment do to the driving frequency?
Answer:
The pedestrians began as an essentially random forcing, which would have produced small uncorrelated motion. But once a slight sway began, people adjusted their steps to match it, which locked the driving frequency onto the natural frequency — turning random forcing into resonance.
The amplitude then grew, which made more people adjust, which drove it harder. The feedback is what made the effect so sudden.
Stiffening the bridge would have raised the natural frequency, which might have moved it away from walking pace but would have been enormously expensive and might have landed on another problematic frequency. Adding dampers instead changes the case from resonant to damped: energy fed in by the crowd is removed faster than it accumulates, so the amplitude settles rather than growing. Thirty-seven viscous dampers and fifty-two tuned mass dampers were fitted, and the bridge reopened.
The engineering lesson is the one this lesson makes algebraically: resonance is unbounded and damping is what bounds it, so a system that cannot avoid its resonant frequency must dissipate energy instead.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Does pulling a mass further down before releasing it change how long each bounce takes?
Correct: No, the period depends only on the mass and the spring.
\[ \omega = \sqrt{\frac{k}{m}} \quad \text{contains no } x_0 \text{ and no } v_0 \]
Why: Omega is the square root of k over m and contains neither initial condition, so the period is unaffected by how the motion was started. Physically, a larger displacement means a proportionally larger restoring force, so the object moves faster over the longer distance and arrives in the same time. This independence is what makes a pendulum clock keep time as its swing decays, and it is the single most surprising fact about simple harmonic motion.
Explain it
They accept that a Ferris wheel gives a sinusoid because it goes round. They cannot see why a bouncing spring should.
Discussion prompt
In no more than five sentences, explain why the spring's motion is sinusoidal even though nothing rotates.
Hint: What is the same about the two situations, if it is not the circle?
Answer:
A usable answer: forget the circle for a moment and ask what makes a Ferris wheel's height a sine wave. It is that the further from the middle you are, the harder something pulls you back — and the pull is proportional to how far out you are.
A spring does exactly that: stretch it twice as far and it pulls back twice as hard, which is Hooke's law. Anything obeying that rule oscillates sinusoidally, whether or not it travels in a circle. The circle was one example of the rule rather than the reason for it, and this is why sinusoids describe so many things that plainly do not go round — sound, current, tides, light.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The weight-to-mass conversion is fixed by checking the units of the answer: omega must come out as one over time. The phase is fixed by writing both conditions down before solving either, so the quadrant is determined rather than guessed. Collapsing is fixed by factoring out the common part first and only then applying the sinusoid method. The period of a sum is fixed by taking the frequency ratio in lowest terms and checking that both quotients come out whole. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw a spring with a mass on it in three states — unstretched, at equilibrium, and displaced — and write beside it the sign convention for displacement and velocity, and Hooke's law. Underneath, write the equation for free undamped harmonic motion with all three formulas, and draw a line separating the inputs that determine omega from those that determine A and phi. In the middle, work a complete example: a 32 pound object stretching a spring 2 feet, released 4 feet below equilibrium moving upward at 6 feet per second, finding omega, A and the two phase conditions. On the right, draw three small graphs side by side for free, damped and resonant motion, with their envelopes dashed, and write the amplitude function under each. In the bottom right, sketch a beat pattern and write both the sum form and the product form, marking which factor is the envelope. Finally, circle the one quantity on the page that is unaffected by how the motion was started, and write one sentence saying why that matters.
The circled quantity is omega, and it matters because it means the period of an oscillation can be predicted from the apparatus alone, before anything is set moving — which is what makes a pendulum clock, a tuning fork and a quartz oscillator possible.
Recap
Five things, and the second is the fact that makes oscillation useful for timekeeping.
| If the question says | Your first move is |
|---|---|
| An object weighing so many pounds | Divide by g to get the mass |
| A spring stretches so far | Hooke's law gives the spring constant |
| Released from rest | The amplitude is the initial displacement |
| Released with a velocity | Use the full amplitude formula |
| A sum of two sinusoids | Take the frequency ratio in lowest terms |
Sinusoids are finished. The next two lessons leave them entirely and return to triangles — but this time triangles with no right angle, which neither the unit circle nor the right-triangle ratios can handle on their own.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 885-891 — everything on these slides traces back here
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