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One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120 \(\mathrm{Hz}\) . The other end passes over a pulley and supports a \(1.50-\mathrm{kg}\) mass. The linear mass density of the rope is 0.0550 \(\mathrm{kg} / \mathrm{m}\) . (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 \(\mathrm{kg}\) ?

Short Answer

Expert verified
(a) 16.37 m/s, (b) 0.1364 m, (c) With 3.00 kg mass: Speed = 23.16 m/s, Wavelength = 0.193 m.

Step by step solution

01

Identify given values (Part a & b)

Start by identifying the given values for parts (a) and (b):- Frequency of rope vibration, \( f = 120 \, \text{Hz} \).- Mass supported by the rope, \( m = 1.50 \, \text{kg} \).- Linear mass density of the rope, \( \mu = 0.0550 \, \text{kg/m} \).
02

Calculate tension in the rope due to mass (Part a & b)

The tension (\( T \)) in the rope can be calculated using the formula: \[ T = m \cdot g \] where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Substitute \( m = 1.50 \, \text{kg} \) to find:\[ T = 1.50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 14.715 \, \text{N} \]
03

Find speed of the wave (Part a)

The speed of a transverse wave on the rope, \( v \), is given by the formula: \[ v = \sqrt{\frac{T}{\mu}} \]Substitute the tension (\( T = 14.715 \, \text{N} \)) and linear mass density (\( \mu = 0.0550 \, \text{kg/m} \)) to calculate:\[ v = \sqrt{\frac{14.715}{0.0550}} \approx 16.37 \, \text{m/s} \]
04

Calculate wavelength of the wave (Part b)

The wavelength \( \lambda \) of the wave can be calculated using the wave speed formula: \[ \lambda = \frac{v}{f} \]Substitute the speed of the wave (\( v = 16.37 \, \text{m/s} \)) and the frequency (\( f = 120 \, \text{Hz} \)) to get:\[ \lambda = \frac{16.37}{120} \approx 0.1364 \, \text{m} \]
05

Recalculate for increased mass (Part c)

Repeat Steps 2 to 4 using the increased mass \( m = 3.00 \, \text{kg} \):- New tension: \[ T = 3.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 29.43 \, \text{N} \]- New wave speed: \[ v = \sqrt{\frac{29.43}{0.0550}} \approx 23.16 \, \text{m/s} \]- New wavelength: \[ \lambda = \frac{23.16}{120} \approx 0.193 \, \text{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transverse Waves
Transverse waves are a type of wave where the movement of the rope is perpendicular to the direction of the wave's propagation. Imagine a wave traveling along a rope; as the wave passes, the particles of the rope move up and down, but the wave itself moves horizontally across.
This is different from longitudinal waves, where the particle movement is in the same direction as the wave.

Transverse waves are common in many scenarios, such as in light waves and on strings. The vibrations we encounter in the problem are especially important to understand as they demonstrate a direct application of wave properties in physics. A tuning fork creates such waves on a string, making it ideal for studying these dynamics closely.
Wave Speed Calculation
The speed of a wave on a string is a crucial concept in wave mechanics. It gives insight into how fast a wave travels along a medium. In our exercise, we calculate the wave speed using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density. The wave speed describes how energy and information pass along the string rapidly, so understanding how to compute it is vital.
  • When the tension increases, the speed of the wave increases since the medium (rope) becomes tighter, resulting in a stiffer connection between molecules.
  • Alternatively, if the linear mass density increases, the wave speed decreases as the medium becomes heavier, slowing down the wave's propagation.
The formula also shows how the nature of the medium affects the wave speed, illustrating how physical changes influence wave characteristics.
Linear Mass Density
Linear mass density \( \mu \) is an essential property of the rope that determines how mass is distributed along its length. This quantity helps in understanding how the medium can affect the behaviour of a wave traveling through it. Its value is defined as the mass per unit length of the rope:
\[ \mu = \frac{\text{mass of the rope}}{\text{length of the rope}} \]

In our exercise, the linear mass density is given as \( 0.0550 \, \text{kg/m} \). A higher density means more material is present in the same length of the string, influencing wave speed and tension.
  • Linear mass density is critical in calculations for wave speed as it directly appears in the denominator, affecting the speed inversely.
  • In practical terms, knowing the linear mass density helps in designing instruments and systems where precise wave movement is required, such as musical instruments or engineering applications.
Tension in Strings
Tension plays a significant role in determining the properties of waves on a string. Tension arises due to forces applied to the string, resulting in stretches that create the potential energy needed for wave propagation. In this exercise, tension is calculated using the mass added to the rope and gravity, as per the formula:
\[ T = m \cdot g \]
where \( m \) is the mass suspended by the string and \( g = 9.81 \, \text{m/s}^2 \) is the gravitational pull.

In our problem, increasing the mass increases the tension.
  • A higher tension results in a string that's more resistant to bending, which affects how quickly a wave can travel through the medium.
  • Conversely, less tension results in slower waves due to the reduced force holding the particles together.
This principle is key to understanding string instruments or calculating safe load limits on cables and wires in engineering. Knowing how to control and calculate tension allows for fine-tuning of wave behaviors and practical applications.

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Most popular questions from this chapter

15.44\(\cdot\) The wave function of a standing wave is \(y(x, t)\) 4.44 \(\mathrm{mm} \sin [(32.5 \mathrm{rad} / \mathrm{m}) x] \sin [(754 \mathrm{rad} / \mathrm{s}) t] .\) For the two traveing waves that make up this standing wave, find the (a) amplitude (b) wavelength; (c) frequency; (d) wave speed; (e) wave function (f) From the information given, can you determine which hamonic this is? Explain.

With what tension must a rope with length 2.50 \(\mathrm{m}\) and mass 0.120 \(\mathrm{kg}\) be stretched for transverse waves of frequency 40.0 \(\mathrm{Hz}\) to have a wavelength of 0.750 \(\mathrm{m} ?\)

A transverse sine wave with an amplitude of 2.50 \(\mathrm{mm}\) and a wavelength of 1.80 \(\mathrm{m}\) travels from left to right along a long, horizontal, stretched string with a speed of 36.0 \(\mathrm{m} / \mathrm{s} .\) Take the origin at the left end of the undisturbed string. At time \(t=0\) the left end of the string has its maximum upward displacement. (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\) for a particle 1.35 \(\mathrm{m}\) to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverss velocity of a particle 1.35 \(\mathrm{m}\) to the right of the origin at time \(t=0.0625 \mathrm{s}\)

CALC Ant Joy Ride. You place your pet ant Klyde (mass \(m )\) on top of a horizontal, stretched rope, where he holds on tightly. The rope has mass \(M\) and length \(L\) and is under tension \(F .\) You start a sinusoidal transverse wave of wavelength \(\lambda\) and amplitude \(A\) propagating along the rope. The motion of the rope is in a vertical plane. Klyde's mass is so small that his presence has no effect on the propagation of the wave. (a) What is Klyde's top speed as he oscillates up and down? (b) Klyde enjoys the ride and begs for more. You decide to double his top speed by changing the tension while keeping the wavelength and amplitude the same. Should the tension be increased or decreased, and by what factor?

Tuning an Instrument. A musician tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g} .\) (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 \(\mathrm{Hz}\) to 73.4 \(\mathrm{Hz}\) , corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

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