Reading a sine wave
Tonight's idea
The plain wave \(y = \sin x\), with \(x\) in radians, starts at 0, swings between \(-1\) and 1, and has period \(2\pi\): it repeats that often. Every other sine wave is that one after a shift, stretch or flip. Four numbers say which.
\(y = a\sin\big(b(x - c)\big) + d\)
- \(d\) lifts. The wave swings about the midline \(y = d\), halfway between its highest and lowest values.
- \(a\) sets the height. The amplitude \(|a|\) is the distance from the midline to a peak: half of highest minus lowest.
- \(b\) sets the speed. The bracket must grow by \(2\pi\) for one full wave, so \(x\) must grow by \(2\pi/b\). That length is the period, the gap from one peak to the next.
- \(c\) slides. The plain wave crosses its midline going up at 0. This one does so when the bracket is zero, at \(x = c\).
Watch one: the wave in the figure
Highest value 5, lowest 1. The midline is halfway: \(d = \tfrac{5 + 1}{2} = 3\). The amplitude is half the gap: \(a = \tfrac{5 - 1}{2} = 2\).
Peaks sit at \(x = 2\) and \(x = 6\), so the period is 4. Then \(\tfrac{2\pi}{b} = 4\) gives \(b = \tfrac{\pi}{2}\).
The wave crosses the midline going up at \(x = 1\), so \(c = 1\).
\[y = 2\sin\!\Big(\frac{\pi}{2}(x - 1)\Big) + 3.\]
Check a peak: at \(x = 2\) the bracket is \(\tfrac{\pi}{2}\), where sine is 1, and \(y = 2 + 3 = 5\).
Move the four sliders and match a mystery wave: mathdetours.com/demos/sinewave.html.
Your turn
- For \(y = 3\sin\big(2(x - \tfrac{\pi}{4})\big) - 1\), give the amplitude, the period, the midline, and the highest and lowest values.
- A sine wave has highest value 7, lowest value 1 and period \(\pi\). It crosses its midline going up at \(x = 0\). Write its equation.
- A wave has a peak at \((2, 9)\) and its next trough at \((5, -1)\). Find \(a\), \(d\), the period and \(b\). Then work out where it crosses the midline going up, and write an equation.
- Sam says \(y = \sin(2x - \tfrac{\pi}{3})\) is the wave \(y = \sin 2x\) slid right by \(\tfrac{\pi}{3}\). Find where the wave really crosses its midline going up, and rewrite the equation so the slide can be read off. Why was Sam out by a factor of 2?
Stretch: one wave, many names
The wave in the figure has other equations. Find one of the form \(y = -2\sin\big(\tfrac{\pi}{2}(x - c)\big) + 3\), and one of the form \(y = 2\cos\big(\tfrac{\pi}{2}(x - c)\big) + 3\). What does each \(c\) mark on the graph?
Answers
- Amplitude 3, period \(\tfrac{2\pi}{2} = \pi\), midline \(y = -1\). Highest value \(-1 + 3 = 2\), lowest \(-1 - 3 = -4\).
- \(d = \tfrac{7 + 1}{2} = 4\), \(a = \tfrac{7 - 1}{2} = 3\), \(b = \tfrac{2\pi}{\pi} = 2\), \(c = 0\). So \(y = 3\sin 2x + 4\).
- \(d = \tfrac{9 + (-1)}{2} = 4\) and \(a = \tfrac{9 - (-1)}{2} = 5\). Peak to trough is half a period, so the period is 6 and \(b = \tfrac{2\pi}{6} = \tfrac{\pi}{3}\). The upward crossing is a quarter period before the peak: \(2 - 1.5 = 0.5\). So \(y = 5\sin\big(\tfrac{\pi}{3}(x - 0.5)\big) + 4\).
- The upward crossing is where the bracket is zero: \(2x - \tfrac{\pi}{3} = 0\), so \(x = \tfrac{\pi}{6}\). Factor out the 2: \(y = \sin\big(2(x - \tfrac{\pi}{6})\big)\), a slide of \(\tfrac{\pi}{6}\). The 2 makes the wave run twice as fast, so a change of \(\tfrac{\pi}{3}\) inside the bracket is used up by a change of only \(\tfrac{\pi}{6}\) in \(x\).
Stretch. A negative \(a\) flips the wave, so \(c\) must mark a downward crossing: \(c = 3\), giving \(y = -2\sin\big(\tfrac{\pi}{2}(x - 3)\big) + 3\). Cosine starts at its highest value, so \(c\) must mark a peak: \(c = 2\), giving \(y = 2\cos\big(\tfrac{\pi}{2}(x - 2)\big) + 3\). Adding any whole number of periods to \(c\) also works (\(c = 7\), or \(c = 6\)).