Unformatted text preview: .55 = 12.55 − 3.3 = 9.25. At this point, we have
H (t) = 9.25 sin(ωt + φ) + 12.55. Next, we go after the angular frequency ω . Since the data
collected is over the span of a year (12 months), we take the period T = 12 months.6 This
5 Okay, it appears to be the ‘∧’ shape we saw in some of the graphs in Section 2.2. Just humor us.
Even though the data collected lies in the interval [1, 12], which has a length of 11, we need to think of the data
point at t = 1 as a representative sample of the amount of daylight for every day in January. That is, it represents
H (t) over the interval [0, 1]. Similarly, t = 2 is a sample of H (t) over [1, 2], and so forth.
6 11.1 Applications of Sinusoids 751 means ω = 2π = 2π = π . The last quantity to ﬁnd is the phase φ. Unlike the previous
example, it is easier in this case to ﬁnd the phase shift − ω . Since we picked A > 0, the phase
shift corresponds to the ﬁrst value of t with H (t) = 12.55 (the baseline value).7 Here, we
choose t = 3, since its corresponding H value of 12.4 is close...
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