Choose sine and cosine functions to model periodic phenomena with a given amplitude, period, frequency and midline

Learning goal: Choose trigonometric functions to model periodic phenomena, reading amplitude, frequency, period and midline from a description or a graph, and writing a sine or cosine function with those specified features.

What periodic means. A quantity is periodic when it repeats the same pattern over and over at equal time intervals: the height of a seat on a Ferris wheel, the depth of water at a harbor as the tide rises and falls, the position of a swinging pendulum, the hours of daylight through a year. Sine and cosine are the natural functions for such quantities because they are periodic themselves and because the unit circle, a point going round and round, is the simplest periodic motion there is. A periodic quantity need not follow a sine curve, but when a sinusoidal model fits, the four numbers that describe it are its midline, amplitude, period and frequency, and the modeling task is to read those four numbers from the situation and build the function that has them.

The four features of y = A sin(B(t − C)) + D. Take the basic sine function and stretch and shift it. The midline is the horizontal line y = D halfway between the highest and lowest values; D is the average of the maximum and minimum. The amplitude is A, the distance from the midline to the maximum (or to the minimum); A is half the difference between maximum and minimum, and we take A positive. The period is the time for one complete cycle. The basic sine function has period 2π, and multiplying the input by a positive number B changes that to 2π/B (a horizontal compression when B > 1, a stretch when 0 < B < 1), so period = 2π/B, equivalently B = 2π/period (we take A and B positive throughout). The frequency is the number of cycles per unit of time, which is 1 divided by the period, so frequency = B/(2π). The number C shifts the graph horizontally. The same four features describe y = A cos(B(t − C)) + D; the only difference between sine and cosine here is where the cycle starts: cosine starts at a maximum when C = 0, sine starts on the midline going up.

Choosing sine or cosine. Neither is "correct"; choose the one that makes C simplest. If the data begin at a maximum, use cosine with no shift. If the data begin at a minimum, use −A cos(Bt) + D, which starts at its lowest value. If the data begin on the midline and rise, use sine with no shift. If the known point is neither, use either function and solve for the shift C, which is the time at which that function's basic cycle begins.

A fully worked model: a hypothetical tide. Suppose a harbor (this is an invented example, not a real tide table) has high water 5.2 m deep at 3:00 a.m. and the next low water 1.0 m deep at 9:12 a.m., and suppose the depth is well modeled by a single sinusoid. Measure t in hours after midnight, so high water is at t = 3 and low water at t = 9.2. Midline: D = (5.2 + 1.0)/2 = 3.1 m. Amplitude: A = (5.2 − 1.0)/2 = 2.1 m. Period: high to low is half a cycle, 9.2 − 3 = 6.2 hours, so the period is 12.4 hours and B = 2π/12.4 ≈ 0.5067. Frequency: 1/12.4 ≈ 0.0806 cycles per hour. Because the known point is a maximum, choose cosine with its start at t = 3: h(t) = 3.1 + 2.1 cos(2π(t − 3)/12.4). Check: h(3) = 3.1 + 2.1 cos 0 = 5.2 ✓; h(9.2) = 3.1 + 2.1 cos(2π × 6.2/12.4) = 3.1 + 2.1 cos π = 3.1 − 2.1 = 1.0 ✓. Predict the depth at noon: h(12) = 3.1 + 2.1 cos(2π × 9/12.4) = 3.1 + 2.1 cos(4.561) ≈ 3.1 + 2.1(−0.1514) ≈ 2.78 m. The model also predicts the next high water at 3 + 12.4 = 15.4 hours, that is 3:24 p.m.

A second model: a Ferris wheel. A wheel of radius 20 m has its lowest seat 2 m above the ground and turns once every 4 minutes. The midline is the height of the axle, 2 + 20 = 22 m; the amplitude is the radius, 20 m; the period is 4 minutes, so B = 2π/4 = π/2; the frequency is 1/4 = 0.25 revolutions per minute. A rider boards at the bottom at t = 0, a minimum, so h(t) = 22 − 20 cos(πt/2). Check: h(0) = 22 − 20 = 2 m, h(1) = 22 − 20 cos(π/2) = 22 m, h(2) = 22 + 20 = 42 m at the top, h(4) = 2 m again.

Reading features from a graph. Given a graph, find the highest and lowest values first; their average is the midline and half their difference is the amplitude. Then find the time between two consecutive maxima; that is the period, and B = 2π/period. Finally find one point where you know the phase: a maximum for cosine or a midline crossing going up for sine. A common trap is to measure the period from a maximum to the next minimum, which is only half a cycle; another is to confuse the amplitude with the maximum value when the midline is not 0.

The misconception. Students frequently write y = 2.1 sin(12.4t) for a period of 12.4 hours, putting the period where B belongs. That function has period 2π/12.4 ≈ 0.507 hours, a cycle every half hour, which is absurd for a tide. The number multiplying t is not the period; it is 2π divided by the period. Always check the finished model at two known points, a maximum and a minimum, before using it.

Worked, a spring: A weight on a spring bobs between 4 cm below and 4 cm above its rest position, completing 3 full oscillations every 2 seconds, and it starts at the rest position moving upward. Midline 0, amplitude 4 cm, frequency 3/2 = 1.5 cycles per second, period 2/3 s, so B = 2π/(2/3) = 3π. Starting at the midline going up means sine with no shift: y(t) = 4 sin(3πt). Check: y(0) = 0; y(1/6) = 4 sin(π/2) = 4 cm, the top, one quarter of a period after the start; y(2/3) = 4 sin(2π) = 0, back to the start after one full period.

Worked, a hypothetical daylight model: Suppose a town has 12.2 hours of daylight on day 81 of the year, a maximum of 15.3 hours on about day 172, the daylight is increasing on day 81, and the pattern repeats every 365 days (an invented, simplified example). Midline 12.2 h (from the assumption below), amplitude 15.3 − 12.2 = 3.1 h, period 365 days, B = 2π/365. Assume a sinusoidal model, with the daylight changing at a steady phase rate, and assume day 81, when the days are lengthening, is the midline crossing on the way up (its timing, about a quarter year before the maximum, supports that choice); then 12.2 h is the midline, and sine shifted to start at day 81 fits: L(d) = 12.2 + 3.1 sin(2π(d − 81)/365). Check: L(81) = 12.2; L(172) = 12.2 + 3.1 sin(2π × 91/365) = 12.2 + 3.1 sin(1.5665) ≈ 12.2 + 3.1(1.000) = 15.3 ✓ (91/365 of a turn is 0.2493 of a turn, very nearly a quarter). Minimum predicted at day 81 + 273.75 ≈ day 355: L(355) ≈ 9.1 h.

Worked, reading a graph: A graph oscillates between a maximum of 14 and a minimum of 6, with maxima at t = 1 and t = 7. Midline (14 + 6)/2 = 10, amplitude (14 − 6)/2 = 4, period 7 − 1 = 6, so B = 2π/6 = π/3, frequency 1/6. A maximum at t = 1 makes cosine natural: y = 10 + 4 cos(π(t − 1)/3). Check: y(1) = 14; y(4) = 10 + 4 cos(π) = 6, a minimum halfway between the maxima ✓. The same graph is also y = 10 + 4 sin(π(t + 0.5)/3), because sine reaches its maximum a quarter period, 1.5 units, after its midline crossing, and 1 − 1.5 = −0.5.

Representation task: Make a table of the tide model h(t) = 3.1 + 2.1 cos(2π(t − 3)/12.4) every hour from t = 0 to t = 24, using a calculator in radian mode, and plot the 25 points on graph paper with time on the horizontal axis and depth in meters on the vertical axis. Draw the midline y = 3.1, mark the amplitude 2.1 as a vertical arrow from the midline to the maximum, and mark one full period of 12.4 hours from the maximum at t = 3 to the next maximum at t = 15.4. Label the minimum at t = 9.2 and the noon value 2.78 m.

Independent performance: Invent a periodic situation of your own (a rotating wheel, a hypothetical temperature cycle, a bouncing spring) and state a maximum value, a minimum value, the times at which they occur and one further known point. From those numbers compute the midline, amplitude, period and frequency with units, decide whether sine or cosine gives the simplest model and explain why, write the function, and verify it at the maximum, the minimum and the extra point. Then use the model to predict the value at a time not in your data. Success criteria: the four features are computed correctly and carry units; B equals 2π divided by the period; the verification shows the substitutions; the prediction is between the minimum and maximum; and the write-up says clearly that the scenario is invented.