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Write down the amplitude of the wave in metres. Label the points marked:

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5 parts
2.2

40 cm\text{40 cm}

2.4

0.8 m\text{0.8 m}

C and E\text{C and E}

2.6

6 s\text{6 s}

2.7

0.4 m⋅s−1\text{0.4 m} \cdot \text{s}^{-1}

3.1

0.40 m\text{0.40 m}

Step-by-step solution

6 steps

Here are the solutions for Question 2 and Question 3.

Part 2

Part 2.1

A water wave is a *transverse wave

Part 2.2

40 cm\text{40 cm}

From the graph, the maximum displacement from the equilibrium position is 40 cm40 \text{ cm}.

The amplitude of the wave is 40 cm\boxed{\text{40 cm}}.

2.3 Wavelength (λ\lambda) is defined as the distance between two consecutive points in phase on a wave. This could be the distance between two successive crests, two successive troughs, or any two corresponding points on consecutive waves.

Part 2.4

0.8 m\text{0.8 m}

Assumption: The speed and frequency provided in the preamble for Question 3 (v=0.4 m⋅s−1v = 0.4 \text{ m} \cdot \text{s}^{-1} and f=0.5 Hzf = 0.5 \text{ Hz}) apply to this question.

  1. Recall the wave speed formula.

    v=fλv = f \lambda

    Where vv is the wave speed, ff is the frequency, and λ\lambda is the wavelength.

  2. Rearrange the formula to solve for wavelength and substitute the given values.

    λ=vf\lambda = \frac{v}{f} λ=0.4 m⋅s−10.5 Hz\lambda = \frac{0.4 \text{ m} \cdot \text{s}^{-1}}{0.5 \text{ Hz}} λ=0.8 m\lambda = 0.8 \text{ m}

    The wavelength of the wave is 0.8 m\boxed{\text{0.8 m}}.

    2.5 Points in phase are points on the wave that have the same displacement and are moving in the same direction. From the wave form, points C and E are both crests, meaning they are at their maximum positive displacement and are in the same phase.
    Two points on the wave form that are in phase are C and E\boxed{\text{C and E}}.

Part 2.6

6 s\text{6 s}

Assumption: The frequency provided in the preamble for Question 3 (f=0.5 Hzf = 0.5 \text{ Hz}) applies to this question.

  1. Calculate the period (TT) of the wave.

    The period is the inverse of the frequency.

    T=1fT = \frac{1}{f} T=10.5 HzT = \frac{1}{0.5 \text{ Hz}} T=2 sT = 2 \text{ s}

  2. Determine the time taken for FOUR crests to move past a certain point.

    The time between the first crest and the fourth crest passing a point is equivalent to three periods (1st to 2nd is 1T, 2nd to 3rd is 1T, 3rd to 4th is 1T).

    Time=3×T\text{Time} = 3 \times T Time=3×2 s\text{Time} = 3 \times 2 \text{ s} Time=6 s\text{Time} = 6 \text{ s}

    The time taken for four crests to move past a certain point is 6 s\boxed{\text{6 s}}.

Part 2.7

0.4 m⋅s−1\text{0.4 m} \cdot \text{s}^{-1}

Assumption: The frequency provided in the preamble for Question 3 (f=0.5 Hzf = 0.5 \text{ Hz}) applies to this question, and the wavelength calculated in 2.4 (λ=0.8 m\lambda = 0.8 \text{ m}) is used.

  1. Recall the wave speed formula.

    v=fλv = f \lambda

  2. Substitute the frequency and wavelength values.

    v=(0.5 Hz)×(0.8 m)v = (0.5 \text{ Hz}) \times (0.8 \text{ m}) v=0.4 m⋅s−1v = 0.4 \text{ m} \cdot \text{s}^{-1}

    The speed of the wave is 0.4 m⋅s−1\boxed{\text{0.4 m} \cdot \text{s}^{-1}}.

Part 3

Part 3.1

0.40 m\text{0.40 m}

From the graph, the maximum amplitude is 40 cm40 \text{ cm}. To convert this to metres, divide by 100.

40 cm=40100 m=0.40 m40 \text{ cm} = \frac{40}{100} \text{ m} = 0.40 \text{ m}

The amplitude of the wave in metres is 0.40 m\boxed{\text{0.40 m}}.

Part 3.2

  • Point A: Node (or equilibrium position)
  • Point B: Node (or equilibrium position)
  • Point C: Crest
  • Point D: Trough
  • Point E: Crest
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