Read harmonic parameters and complete cycles · Worked example
A laboratory signal is modeled by \(y(t)=-5\cos\left(0.35t-1.570796326795\right)\), with time in seconds and output in centimeters. Find amplitude, period, frequency and midline, then sketch one complete cycle and tabulate five quarter-cycle points. Explain whether the coefficient sign changes amplitude or only orientation. Show the derivation, labeled graph and contextual reasoning. Written reasoning and sketches require teacher review; selecting a summary does not verify them. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
Answer: Amplitude: \(5\); Period (s): \(17.952\); Frequency (Hz): \(0.0557\); Midline: \(0\).
Amplitude is \(|-5|=5\), never negative. The midline is \(y=0\). Angular frequency is \(\omega=0.35\) radians per second. Thus \(P=\dfrac{2\pi}{\omega}=17.9519580205\) seconds and \(f=\dfrac{1}{P}=0.05570423\) hertz. For very short periods use full precision, not a display rounded to zero. The argument completes \(2\pi\) radians over one cycle. Phase and a negative scale alter the starting position or orientation, not the positive period.
Five quarter-cycle points, rounded only for display
Time
Value
\(4.488\)
\(-5\)
\(8.976\)
\(0\)
\(13.464\)
\(5\)
\(17.952\)
\(0\)
\(22.4399\)
\(-5\)
One complete cycle, with all phase and baseline shifts Written-work review:
Use every stated input, sign convention, time origin and unit; derive each requested quantity.
Draw all requested curves and include all observed points and bounded interval endpoints.
Accept equivalent exact forms and numerical results within one unit in the last displayed decimal, unless a stricter tolerance is stated. Written proofs, interpretation and sketches require teacher review.