The diagram below shows a portion a plastic fibre cable, which allows light waves to pass through. The path of the light wave can be modelled by a trigonometric function.

  1. (a)

    It is given that the diameter of the cable is 4 millimetres.
    Find the amplitude of the light wave.

  2. (b)

    It is given further that the period of the light wave is 500 nanometres.
    Find the length of the portion of cable shown in the diagram.

  3. (c)

    Which of the following can be a suitable model for the light wave?

    y=2sin(πx)y = 2 sin(\pi x)

    y=2cos⁡(πx)y = 2\cos(\pi x)

    y=2sin⁡(π250x)y = 2\sin(\frac{\pi}{250}x)

    Explain your answer.

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Worked solution

11(a)

1. Determine the amplitude from the diameter of the cable

Amplitude=Diameter2=42=2 mm\text{Amplitude} = \frac{\text{Diameter}}{2} = \frac{4}{2} = 2 \text{ mm}

Answer

22 mm

11(b)

1. Identify the number of cycles in the diagram and calculate the total length

Number of cycles=3\text{Number of cycles} = 3
Length=3×500=1500 nm\text{Length} = 3 \times 500 = 1500\text{ nm}

Answer

1500 nm1500\text{ nm}

11(c)

1. Identify the amplitude and y-intercept to eliminate options

From part (a), the amplitude is 22. The graph starts at (0,0)(0,0), so when x=0x=0, y=0y=0.
For y=2cos⁡(πx)y = 2\cos(\pi x), when x=0x=0, y=2cos⁡(0)=2≠0y = 2\cos(0) = 2 \neq 0. Thus, it is not a cosine function.

2. Use the period to determine the coefficient of x

From part (b), the period is 500500 nm.
For a sine function y=asin⁡(bx)y = a\sin(bx), the period is 2πb\frac{2\pi}{b}.
2πb=500  ⟹  b=2π500=π250\frac{2\pi}{b} = 500 \implies b = \frac{2\pi}{500} = \frac{\pi}{250}.

3. Conclude the correct model

Therefore, the suitable model is y=2sin⁡(π250x)y = 2\sin\left(\frac{\pi}{250}x\right).

Answer

y=2sin⁡(π250x)y = 2\sin\left(\frac{\pi}{250}x\right)

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