Question
Two sections of rope have waves traveling on them. In rope 1, the waves appear very close together. In rope 2, the waves are very spaced out. Which physical properties can be determined about these waves?(1 point) Responses Rope 1 has a higher frequency and a longer wavelength than rope 2. Rope 1 has a higher frequency and a longer wavelength than rope 2. Rope 1 has a lower frequency and a longer wavelength than rope 2. Rope 1 has a lower frequency and a longer wavelength than rope 2. Rope 1 has a lower frequency and a shorter wavelength than rope 2. Rope 1 has a lower frequency and a shorter wavelength than rope 2. Rope 1 has a higher frequency and a shorter wavelength than rope 2
Answers
GPT-4o mini
To determine the relationship between the properties of the waves on the two ropes, we can analyze the information given:
- Waves that are "very close together" indicate a higher frequency. This is because frequency (f) is the number of waves that pass a given point per unit time.
- Waves that are "very spaced out" indicate a lower frequency.
Additionally, the wavelength (λ) is the distance between successive crests of a wave.
- Since wave speed (v) is related to frequency and wavelength by the equation \( v = f \cdot \lambda \), if one wave has a higher frequency, it typically has a shorter wavelength, assuming wave speed in both ropes is the same.
From the descriptions:
- Rope 1 (waves close together) has a **higher frequency** and a **shorter wavelength** compared to Rope 2 (waves spaced out).
- Rope 2 (waves spaced out) has a **lower frequency** and a **longer wavelength**.
Therefore, the correct response is:
**Rope 1 has a higher frequency and a shorter wavelength than rope 2.**
- Waves that are "very close together" indicate a higher frequency. This is because frequency (f) is the number of waves that pass a given point per unit time.
- Waves that are "very spaced out" indicate a lower frequency.
Additionally, the wavelength (λ) is the distance between successive crests of a wave.
- Since wave speed (v) is related to frequency and wavelength by the equation \( v = f \cdot \lambda \), if one wave has a higher frequency, it typically has a shorter wavelength, assuming wave speed in both ropes is the same.
From the descriptions:
- Rope 1 (waves close together) has a **higher frequency** and a **shorter wavelength** compared to Rope 2 (waves spaced out).
- Rope 2 (waves spaced out) has a **lower frequency** and a **longer wavelength**.
Therefore, the correct response is:
**Rope 1 has a higher frequency and a shorter wavelength than rope 2.**