Unit 4, lesson 54

Amplitude, Period, and Phase Shift

What each of the four constants in a general sine or cosine function does to the graph, and how to find them from a graph or from data that rise and fall.

  • Theory part 107
  • Practice part 108
  • Graphs of Trigonometric Functions

A weight bouncing on a spring, the depth of water in a harbor and the hours of daylight through the year all rise and fall like a sine wave. None of them swings between −1-1 and 11 every 2π2\pi units, though. To describe them you stretch, squeeze and slide the basic wave, and four constants are enough to do all of it.

Key terms

General sinusoidal form

Definition A sinusoidal function is a function of the form

y=Asin⁡(B(x−C))+Dory=Acos⁡(B(x−C))+D,\begin{gathered} y = A\sin\bigl(B(x - C)\bigr) + D \\[0.4em] \text{or} \\[0.4em] y = A\cos\bigl(B(x - C)\bigr) + D, \end{gathered}

where AA, BB, CC and DD are real constants with A≠0A \ne 0 and B≠0B \ne 0, and the input B(x−C)B(x - C) of sine or cosine is measured in radians. When B>0B > 0 the graph has amplitude ∣A∣\lvert A \rvert, period 2πB\frac{2\pi}{B}, horizontal shift CC and midline y=Dy = D, and the range is [D−∣A∣,D+∣A∣][D - \lvert A \rvert, D + \lvert A \rvert]. For the sine form with A>0A > 0, one cycle starts on the midline at x=Cx = C, heading up, and ends at x=C+2πBx = C + \frac{2\pi}{B}.

In plain words Each constant has one job. The number AA sets the size of the swing, BB sets how quickly the cycles come, CC slides the wave left or right, and DD lifts or lowers it.

Example Take the function

y=3sin⁡(2(x−π4))+1.y = 3\sin\left(2\left(x - \frac{\pi}{4}\right)\right) + 1.

Its amplitude is 33, its period is 2π2=π\frac{2\pi}{2} = \pi, it is shifted π4\frac{\pi}{4} units to the right, and its midline is y=1y = 1, so its values fill [−2,4][-2, 4]. One cycle runs from x=π4x = \frac{\pi}{4} to x=5π4x = \frac{5\pi}{4}, and the quarter points of that cycle are the key points.

xxπ4\frac{\pi}{4}π2\frac{\pi}{2}3π4\frac{3\pi}{4}π\pi5π4\frac{5\pi}{4}
yy114411−2-211

Amplitude

Definition The amplitude of y=Asin⁡(B(x−C))+Dy = A\sin\bigl(B(x - C)\bigr) + D, or of the matching cosine function, is ∣A∣\lvert A \rvert. It equals M−m2\frac{M - m}{2}, where M=D+∣A∣M = D + \lvert A \rvert is the maximum value and m=D−∣A∣m = D - \lvert A \rvert is the minimum value. Multiplying by AA stretches the basic wave vertically by the factor ∣A∣\lvert A \rvert (a compression when ∣A∣<1\lvert A \rvert < 1) and also reflects it when A<0A < 0.

In plain words Amplitude is how far the graph rises above its midline and how far it sinks below it. Only the size of AA matters here, not its sign.

Example The graph of y=−4sin⁡xy = -4\sin x has amplitude 44. It reaches 44 at x=3π2x = \frac{3\pi}{2} and −4-4 at x=π2x = \frac{\pi}{2}, the reverse of y=4sin⁡xy = 4\sin x. The graph of y=12cos⁡xy = \frac{1}{2}\cos x has amplitude 12\frac{1}{2}, so it is a flattened cosine wave.

Midline

Definition The midline of a sinusoidal function is the line y=Dy = D. It is the horizontal line halfway between the maximum value MM and the minimum value mm, so

D=M+m2.D = \frac{M + m}{2}.

In plain words The constant DD moves the whole wave up or down without changing its shape, and the midline moves with it.

Example The function y=2cos⁡x−3y = 2\cos x - 3 has midline y=−3y = -3. Its values run from −3−2=−5-3 - 2 = -5 up to −3+2=−1-3 + 2 = -1, so the whole graph sits below the xx-axis.

Period

Definition For B≠0B \ne 0 the period of a sinusoidal function is

P=2π∣B∣.P = \frac{2\pi}{\lvert B \rvert}.

The reason is that as xx moves across an interval of length 2π∣B∣\frac{2\pi}{\lvert B \rvert}, the input B(x−C)B(x - C) moves across an interval of length 2π2\pi, which is exactly one cycle of sine or cosine.

In plain words A larger ∣B∣\lvert B \rvert packs more cycles into the same space, so each cycle is shorter. When ∣B∣\lvert B \rvert is between 00 and 11 the cycles are longer than 2π2\pi.

Example For y=cos⁡(πx3)y = \cos\left(\frac{\pi x}{3}\right) the constant is B=π3B = \frac{\pi}{3}, so

P=2ππ/3=6.P = \frac{2\pi}{\pi / 3} = 6.

As xx runs from 00 to 66, the input πx3\frac{\pi x}{3} runs from 00 to 2π2\pi, one full turn. By contrast y=sin⁡(4x)y = \sin(4x) has period 2π4=π2\frac{2\pi}{4} = \frac{\pi}{2} and fits four cycles into every stretch of length 2π2\pi.

Frequency

Definition The frequency of a sinusoidal function is the number of cycles per unit of xx. It is the reciprocal of the period,

frequency=1P=∣B∣2π.\text{frequency} = \frac{1}{P} = \frac{\lvert B \rvert}{2\pi}.

The number ∣B∣\lvert B \rvert itself is called the angular frequency. It measures how many radians the input advances per unit of xx.

In plain words Period says how long one cycle takes. Frequency says how many cycles fit into one unit. Each is the reciprocal of the other.

Example A pure tone can be modeled by y=sin⁡(880πt)y = \sin(880\pi t) with tt in seconds. Its period is

P=2π880π=1440 second,P = \frac{2\pi}{880\pi} = \frac{1}{440} \text{ second,}

so its frequency is 440440 cycles per second.

Horizontal shift

Definition The graph of y=Asin⁡(B(x−C))+Dy = A\sin\bigl(B(x - C)\bigr) + D is the graph of y=Asin⁡(Bx)+Dy = A\sin(Bx) + D shifted CC units to the right when C>0C > 0, and ∣C∣\lvert C \rvert units to the left when C<0C < 0. The number CC is also called the phase shift. The same holds for cosine. When the formula is written as Asin⁡(Bx−c)+DA\sin(Bx - c) + D, factor out BB first,

Bx−c=B(x−cB),Bx - c = B\left(x - \frac{c}{B}\right),

so the shift is cB\frac{c}{B}, not cc.

In plain words The constant CC tells you where the cycle that normally starts at x=0x = 0 starts instead. Read it only after the inside looks like B(x−C)B(x - C), or you will slide the wave the wrong distance.

Example The function y=sin⁡(2x−π3)y = \sin\left(2x - \frac{\pi}{3}\right) has the inside

2x−π3=2(x−π6),2x - \frac{\pi}{3} = 2\left(x - \frac{\pi}{6}\right),

so the shift is π6\frac{\pi}{6} to the right, not π3\frac{\pi}{3}. A cycle starts at x=π6x = \frac{\pi}{6}, where y=sin⁡0=0y = \sin 0 = 0. It peaks at x=5π12x = \frac{5\pi}{12}, a quarter period later, and ends at x=π6+π=7π6x = \frac{\pi}{6} + \pi = \frac{7\pi}{6}.

Positive and negative parameters

Definition In y=Asin⁡(B(x−C))+Dy = A\sin\bigl(B(x - C)\bigr) + D and in the matching cosine function, the sign of each constant has a fixed meaning.

  • If A<0A < 0, the graph is the graph with the positive constant −A-A in place of AA, reflected across the midline y=Dy = D. The amplitude is still ∣A∣\lvert A \rvert.
  • If B<0B < 0, the function can be rewritten with the positive constant −B-B, as shown below. The period is 2π∣B∣\frac{2\pi}{\lvert B \rvert} either way.
  • If C>0C > 0, the shift is to the right, and if C<0C < 0, it is to the left. For example sin⁡(x+π4)\sin\left(x + \frac{\pi}{4}\right) has C=−π4C = -\frac{\pi}{4}.
  • If D>0D > 0, the midline lies above the xx-axis, and if D<0D < 0, it lies below.

A negative BB can always be made positive because sine is odd and cosine is even,

sin⁡(B(x−C))=−sin⁡(−B(x−C)),cos⁡(B(x−C))=cos⁡(−B(x−C)).\begin{aligned} &\sin\bigl(B(x - C)\bigr) \\[0.4em] &\quad = -\sin\bigl(-B(x - C)\bigr), \\[0.8em] &\cos\bigl(B(x - C)\bigr) \\[0.4em] &\quad = \cos\bigl(-B(x - C)\bigr). \end{aligned}

In plain words A negative AA turns the wave upside down. The minus sign of a negative BB moves out front for sine and simply disappears for cosine. Once the inside is written as B(x−C)B(x - C), a plus sign as in x+π4x + \frac{\pi}{4} means the wave moves left. A negative DD moves it down.

Example Since sine is odd,

y=3sin⁡(−2x)−1=−3sin⁡(2x)−1.\begin{aligned} y &= 3\sin(-2x) - 1 \\[0.4em] &= -3\sin(2x) - 1. \end{aligned}

The graph has amplitude 33, period π\pi and midline y=−1y = -1, and it is upside down compared with y=3sin⁡(2x)−1y = 3\sin(2x) - 1. At x=π4x = \frac{\pi}{4} it reaches its lowest value −4-4, exactly where 3sin⁡(2x)−13\sin(2x) - 1 reaches its highest value 22.

Fitting a sinusoidal model

Definition Fitting a sinusoidal model to a quantity that rises and falls regularly means choosing the constants in

y=Acos⁡(B(x−C))+Dy = A\cos\bigl(B(x - C)\bigr) + D

so that the model has the same maximum value MM, minimum value mm and period PP as the data, and peaks at a time x0x_0 where the data peak. With A>0A > 0 and B>0B > 0 this gives

A=M−m2,D=M+m2,B=2πP,C=x0.\begin{gathered} A = \frac{M - m}{2}, \qquad D = \frac{M + m}{2}, \\[0.8em] B = \frac{2\pi}{P}, \qquad C = x_0. \end{gathered}

The period is the time between two consecutive peaks, or twice the time from a peak to the next low point. A sine model works too, with CC at a time where the data cross the midline going up.

In plain words Read off four facts. These are the highest value, the lowest value, the length of one cycle and the time of one peak. Each fact fixes one constant.

Example In a harbor the water is 9 meters deep at high tide, 2 hours after midnight, and 3 meters deep at the next low tide, 8 hours after midnight. Measure tt in hours after midnight. High tide to low tide is half a cycle, so P=2⋅6=12P = 2 \cdot 6 = 12 hours, and

A=9−32=3,D=9+32=6,B=2π12=π6,C=2.\begin{aligned} A &= \frac{9 - 3}{2} = 3, \\[0.8em] D &= \frac{9 + 3}{2} = 6, \\[0.8em] B &= \frac{2\pi}{12} = \frac{\pi}{6}, \\[0.8em] C &= 2. \end{aligned}

The model, in meters, is

d(t)=3cos⁡(π6(t−2))+6.d(t) = 3\cos\left(\frac{\pi}{6}(t - 2)\right) + 6.

As a check, d(8)=3cos⁡π+6=3d(8) = 3\cos \pi + 6 = 3. The same curve can be written as

d(t)=3sin⁡(π6(t+1))+6,d(t) = 3\sin\left(\frac{\pi}{6}(t + 1)\right) + 6,

because a sine wave peaks a quarter cycle, here 3 hours, after it crosses its midline going up. The fit uses one high tide and one low tide, so it is only as reliable as the assumption that the tides repeat exactly every 12 hours.

Check yourself

  1. Find the amplitude, period, horizontal shift and midline of y=−4cos⁡(3x+π2)−1y = -4\cos\left(3x + \frac{\pi}{2}\right) - 1, and give its range.
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Factoring the inside gives

3x+π2=3(x+π6),3x + \frac{\pi}{2} = 3\left(x + \frac{\pi}{6}\right),

so A=−4A = -4, B=3B = 3, C=−π6C = -\frac{\pi}{6} and D=−1D = -1. The amplitude is 44, the period is 2π3\frac{2\pi}{3}, the graph is shifted π6\frac{\pi}{6} to the left, the midline is y=−1y = -1 and the range is [−5,3][-5, 3].

  1. Write a sine function with amplitude 22, period π\pi, midline y=1y = 1 and a shift of π4\frac{\pi}{4} to the right, using a positive AA. What is its value at x=π2x = \frac{\pi}{2}?
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The period gives B=2ππ=2B = \frac{2\pi}{\pi} = 2, so the function is

y=2sin⁡(2(x−π4))+1.y = 2\sin\left(2\left(x - \frac{\pi}{4}\right)\right) + 1.

At x=π2x = \frac{\pi}{2} the inside is 2⋅π4=π22 \cdot \frac{\pi}{4} = \frac{\pi}{2}, so the value is 2⋅1+1=32 \cdot 1 + 1 = 3, the maximum.

  1. In a certain city the daylight lasts 9 hours on the shortest day and 15 hours on the longest day, which is day 172 of a 365-day year. Write a cosine model H(t)H(t) for the hours of daylight on day tt.
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The amplitude is 15−92=3\frac{15 - 9}{2} = 3, the midline is 15+92=12\frac{15 + 9}{2} = 12, the period is 365365 days and the peak is at t=172t = 172. So

H(t)=3cos⁡(2π365(t−172))+12.H(t) = 3\cos\left(\frac{2\pi}{365}(t - 172)\right) + 12.

The model puts the shortest day half a year after the longest, at t=172+182.5=354.5t = 172 + 182.5 = 354.5, so around day 354354 or 355355.

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