One end of the horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates a rope transversely at 250HZ. The other end passes over a pulley and supports at 3 kg mass. The liner mass density of the rope is 0.048 kg/m. What is the speed of a transverse waves on the rope? (m/s)
One end of the horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates a rope transversely at 250HZ. The other end passes over a pulley and supports at 3 kg mass. The liner mass density of the rope is 0.048 kg/m. What is the speed of a transverse waves on the rope? (m/s)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Statement:**
One end of the horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates a rope transversely at 250 Hz. The other end passes over a pulley and supports a 3 kg mass. The linear mass density of the rope is 0.048 kg/m.
**Question:**
What is the speed of transverse waves on the rope? (m/s)
**Solution:**
To determine the speed of transverse waves on the rope, we can use the following formula for wave speed on a rope under tension:
\[ v = \sqrt{\frac{T}{\mu}} \]
where:
- \( v \) is the wave speed,
- \( T \) is the tension in the rope,
- \( \mu \) is the linear mass density of the rope.
**Step-by-step Calculation:**
1. **Calculate the tension in the rope (T):**
Tension (T) is created by the weight of the mass hanging at the end of the rope.
\[ T = mg \]
where:
- \( m \) is the mass (3 kg),
- \( g \) is the acceleration due to gravity (approximated as 9.8 m/s²).
Substitute the values:
\[ T = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 \]
\[ T = 29.4 \, \text{N} \]
2. **Calculate the wave speed (v):**
Use the formula for the wave speed:
\[ v = \sqrt{\frac{T}{\mu}} \]
where:
- \( \mu \) is given as 0.048 kg/m.
Substitute the values:
\[ v = \sqrt{\frac{29.4 \, \text{N}}{0.048 \, \text{kg/m}}} \]
\[ v = \sqrt{612.5 \, \text{m}^2/\text{s}^2} \]
\[ v \approx 24.74 \, \text{m/s} \]
**Answer:**
The speed of transverse waves on the rope is approximately 24.74 m/s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfe80350-6fa1-469d-aae2-95b4dd3aa499%2F63c58156-f267-4952-a482-98e22384f142%2Feoxd1zl_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
One end of the horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates a rope transversely at 250 Hz. The other end passes over a pulley and supports a 3 kg mass. The linear mass density of the rope is 0.048 kg/m.
**Question:**
What is the speed of transverse waves on the rope? (m/s)
**Solution:**
To determine the speed of transverse waves on the rope, we can use the following formula for wave speed on a rope under tension:
\[ v = \sqrt{\frac{T}{\mu}} \]
where:
- \( v \) is the wave speed,
- \( T \) is the tension in the rope,
- \( \mu \) is the linear mass density of the rope.
**Step-by-step Calculation:**
1. **Calculate the tension in the rope (T):**
Tension (T) is created by the weight of the mass hanging at the end of the rope.
\[ T = mg \]
where:
- \( m \) is the mass (3 kg),
- \( g \) is the acceleration due to gravity (approximated as 9.8 m/s²).
Substitute the values:
\[ T = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 \]
\[ T = 29.4 \, \text{N} \]
2. **Calculate the wave speed (v):**
Use the formula for the wave speed:
\[ v = \sqrt{\frac{T}{\mu}} \]
where:
- \( \mu \) is given as 0.048 kg/m.
Substitute the values:
\[ v = \sqrt{\frac{29.4 \, \text{N}}{0.048 \, \text{kg/m}}} \]
\[ v = \sqrt{612.5 \, \text{m}^2/\text{s}^2} \]
\[ v \approx 24.74 \, \text{m/s} \]
**Answer:**
The speed of transverse waves on the rope is approximately 24.74 m/s.
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