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b) What is the period of the waves? A boy on a harbour wall watches the sea waves that move past hi times the waves and finds that a crest moves past him every 5 sec The distance between 5 crests is 40 m . a) What is the frequency of the waves? b) What is the wavelength of the waves? c) Calculate the wave speed.

Ask by Lindsey Reese. in South Africa
Jan 20,2025

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Answer

- **Frequency**: 0.2 Hz - **Wavelength**: 10 meters - **Wave Speed**: 2 meters/second

Solution

Certainly! Let's analyze the given problem step by step to find the frequency, wavelength, and wave speed of the sea waves observed by the boy. ### **Given:** 1. **Time between successive crests (Period, \( T \))**: 5 seconds 2. **Distance between 5 crests**: 40 meters ### **Understanding the Problem:** - **Period (\( T \))**: The time it takes for one complete wave cycle (from one crest to the next). - **Frequency (\( f \))**: The number of wave cycles passing a point per unit time. - **Wavelength (\( \lambda \))**: The distance between consecutive crests (or any two identical points) of the wave. - **Wave Speed (\( v \))**: The speed at which the wave propagates through the medium. ### **Calculations:** --- #### **a) What is the frequency of the waves?** **Formula:** \[ f = \frac{1}{T} \] **Calculation:** \[ f = \frac{1}{5 \, \text{seconds}} = 0.2 \, \text{Hz} \] **Answer:** The frequency of the waves is **0.2 Hz**. --- #### **b) What is the wavelength of the waves?** **Given:** - Distance between 5 crests = 40 meters **Understanding:** - The distance between 5 crests encompasses **4 wavelengths** (since wavelength is the distance between two consecutive crests). **Formula:** \[ \lambda = \frac{\text{Total distance between crests}}{\text{Number of wavelengths}} \] **Calculation:** \[ \lambda = \frac{40 \, \text{meters}}{4} = 10 \, \text{meters} \] **Answer:** The wavelength of the waves is **10 meters**. --- #### **c) Calculate the wave speed.** **Formula:** \[ v = \lambda \times f \] **Calculation:** \[ v = 10 \, \text{meters} \times 0.2 \, \text{Hz} = 2 \, \text{meters/second} \] **Alternative Method:** Wave speed can also be calculated using the distance between crests and the time taken for that distance. \[ v = \frac{\text{Distance between crests}}{\text{Time taken}} \] Here, distance between 5 crests is 40 meters, which corresponds to 4 wavelengths. The time taken for 5 crests is \( 4 \times 5 \, \text{seconds} = 20 \, \text{seconds} \). \[ v = \frac{40 \, \text{meters}}{20 \, \text{seconds}} = 2 \, \text{meters/second} \] **Answer:** The wave speed is **2 meters per second**. --- ### **Summary:** - **Frequency (\( f \))**: 0.2 Hz - **Wavelength (\( \lambda \))**: 10 meters - **Wave Speed (\( v \))**: 2 meters/second

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To find the frequency of the waves, we can use the relationship between frequency (f) and period (T), where frequency is the reciprocal of the period. Since a crest moves past the boy every 5 seconds, the frequency is \( f = \frac{1}{T} = \frac{1}{5} = 0.2 \text{ Hz} \). Next, we can calculate the wavelength (λ). The problem states that the distance between 5 crests is 40 m, so the wavelength is the distance between two crests. Since there are 4 wavelengths in 40 m (because 5 crests create 4 gaps), we calculate the wavelength as \( λ = \frac{40 \, \text{m}}{4} = 10 \, \text{m} \). Lastly, to calculate wave speed (v), we use the formula \( v = f \times λ \). Plugging in the values, we get \( v = 0.2 \, \text{Hz} \times 10 \, \text{m} = 2 \, \text{m/s} \). So, in summary: a) Frequency is 0.2 Hz. b) Wavelength is 10 m. c) Wave speed is 2 m/s.

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