Question

Difficulty: HardGraphs of Trigonometric Functions

The vertical displacement, d(t)d(t) in centimeters, of a particle executing simple harmonic motion is modeled by the trigonometric function d(t)=Asin(B(tC))+Dd(t) = A \sin(B(t - C)) + D, where A>0A > 0, B>0B > 0, and CC represents the smallest non-negative phase shift in seconds. The graph of d(t)d(t) completes one full cycle every 2π3\frac{2\pi}{3} seconds, has a maximum value of 7 cm7\text{ cm} at t=5π18 secondst = \frac{5\pi}{18}\text{ seconds}, and has a minimum value of 3 cm-3\text{ cm}. What is the value of CC?

  1. A
    π18\frac{\pi}{18}
  2. π9\frac{\pi}{9}Answer
  3. C
    2π9\frac{2\pi}{9}
  4. D
    π3\frac{\pi}{3}
  5. E
    5π18\frac{5\pi}{18}

Answer

π9\frac{\pi}{9}
To find the phase shift CC, first determine BB using the period formula Period=2πB\text{Period} = \frac{2\pi}{B}. Since the period is 2π3\frac{2\pi}{3}, B=3B = 3. The function achieves a maximum when its sine argument equals π2+2kπ\frac{\pi}{2} + 2k\pi. Setting 3(5π18C)=π23\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2} yields 5π63C=π2\frac{5\pi}{6} - 3C = \frac{\pi}{2}, which simplifies to 3C=π33C = \frac{\pi}{3}, giving C=π9C = \frac{\pi}{9}.

Step-by-Step Solution

1
Determine the value of BB from the period of the function.
B=3B = 3
The standard period for a sine function is 2π2\pi. Given that the period is 2π3\frac{2\pi}{3}, we set 2πB=2π3\frac{2\pi}{B} = \frac{2\pi}{3}, which yields B=3B = 3.
2
Determine the maximum value equation for the parent sine function.
Argument equals π2\frac{\pi}{2}
The standard sine function sin(θ)\sin(\theta) achieves its first positive maximum at θ=π2\theta = \frac{\pi}{2}. Thus, for d(t)d(t), the maximum occurs when B(tC)=π2B(t - C) = \frac{\pi}{2}.
3
Substitute B=3B = 3 and t=5π18t = \frac{5\pi}{18} into the argument equation and solve for CC.
C=π9C = \frac{\pi}{9}
Substitute the given values: 3(5π18C)=π2    5π63C=π23\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2} \implies \frac{5\pi}{6} - 3C = \frac{\pi}{2}. Subtract 5π6\frac{5\pi}{6} from both sides to get 3C=3π65π6=2π6=π3-3C = \frac{3\pi}{6} - \frac{5\pi}{6} = -\frac{2\pi}{6} = -\frac{\pi}{3}. Dividing by 3-3 gives C=π9C = \frac{\pi}{9}.

Key Concept

Phase shift and parameter identification from graphs of transformed sine functions
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