/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A transverse wave on a rope is g... [FREE SOLUTION] | 91Ó°ÊÓ

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A transverse wave on a rope is given by $$y(x, t)=(0.750 \mathrm{cm}) \cos \pi\left[\left(0.400 \mathrm{cm}^{-1}\right) x+\left(250 \mathrm{s}^{-1}\right) t\right]$$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of \(t : 0.0005\) s, 0.0010 s. (c) Is the wave wave traveling in the or -direction? (d) The mass per unit length of the rope is 0.0500 \(\mathrm{kg} / \mathrm{m} .\) Find the tension. (e) Find the average power of this wave.

Short Answer

Expert verified
Amplitude: 0.750 cm, Period: 0.0251 s, Frequency: 39.89 Hz, Wavelength: 15.71 cm, Speed: 6.277 m/s. Wave: Negative x-direction. Tension: 1.97 N. Average Power: 11.05 W.

Step by step solution

01

Identify Given Equation Components

The wave equation is given as: \[ y(x, t)=(0.750 \, \text{cm}) \cos \pi\left[\left(0.400 \, \text{cm}^{-1}\right) x+\left(250\, \text{s}^{-1}\right) t\right] \]Identify the amplitude (A), wave number (k), and angular frequency (ω) from the equation:\[ A = 0.750 \, \text{cm} \]\[ k = 0.400 \, \text{cm}^{-1} \]\[ \omega = 250 \, \text{s}^{-1} \]
02

Calculate Amplitude and Period

The amplitude (A) is 0.750 cm.The period (T) is calculated from angular frequency:\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{250 \, \text{s}^{-1}} = \frac{2\pi}{250} \approx 0.0251 \, \text{s} \]
03

Calculate Frequency and Wavelength

The frequency (f) is the reciprocal of the period:\[ f = \frac{1}{T} = \frac{1}{0.0251} \approx 39.89 \, \text{Hz} \]The wavelength (λ) is calculated from the wave number:\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.400 \, \text{cm}^{-1}} = 15.71 \, \text{cm} \]
04

Determine Wave Speed

The speed (v) of the wave is the product of frequency and wavelength:\[ v = f \cdot \lambda = 39.89 \, \text{Hz} \times 15.71 \, \text{cm} = 627.7 \, \text{cm/s} \] or 6.277 m/s.
05

Determine Wave Direction

The wave function is \( y(x, t)=A \cos(kx+\omega t) \).Since the equation has a positive angular frequency term \(+\omega t\), the wave is traveling in the negative x-direction.
06

Sketch Wave Shape

Use the given wave function to sketch the wave at \( t = 0.0005 \, \text{s} \) and \( t = 0.0010 \, \text{s} \), noting changes in phase due to time.
07

Calculate Tension in Rope

The speed of a wave on a string is given by:\[ v = \sqrt{\frac{T}{\mu}} \]Given \( v = 6.277 \, \text{m/s} \) and mass per unit length \( \mu = 0.0500 \, \text{kg/m} \), solve for tension (T):\[ T = \mu v^2 = 0.0500 \, \text{kg/m} \times (6.277 \, \text{m/s})^2 \approx 1.97 \, \text{N} \]
08

Calculate Average Power

The average power (P) of the wave is given by:\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]Substitute the values:\[ P = \frac{1}{2} (0.0500 \, \text{kg/m}) (250 \, \text{s}^{-1})^2 (0.0075 \, \text{m})^2 (6.277 \, \text{m/s}) \approx 11.05 \, \text{W} \]

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Key Concepts

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

Wave Equation
The wave equation is a mathematical representation of a wave in motion. For a transverse wave on a rope, the equation typically takes the form: \[ y(x, t) = A \cos(kx + \omega t) \]where:
  • \( y(x, t) \) represents the wave displacement at a point \( x \) and time \( t \).
  • \( A \) is the amplitude of the wave, indicating the maximum displacement from the equilibrium position.
  • \( k \) is the wave number, expressing the number of wave cycles per unit distance, defined as \( k = \frac{2\pi}{\lambda} \).
  • \( \omega \) is the angular frequency, showing how many oscillations occur per unit time, given by \( \omega = 2\pi f \).
Together, these components describe both the spatial and temporal aspects of the wave’s motion.
The formula helps in understanding how waves propagate through space and time, serving as a foundational concept in physics.
Wave Parameters
Wave parameters are essential characteristics that define the behavior of a wave. These include amplitude, period, frequency, and wavelength.
**Amplitude** is the height of the wave peak from its rest position. It represents the energy of the wave, where higher amplitude means more energy. In the given problem, the amplitude is 0.750 cm.The **period** (\( T \)) is the time taken for one complete wave cycle to pass a given point. It's calculated as the reciprocal of frequency: \( T = \frac{1}{f} \).
**Frequency** (\( f \)) is how many wave cycles occur in a second, given in hertz (Hz). It is implicated in sound waves as pitch or color in light waves.
**Wavelength** (\( \lambda \)) is the distance between consecutive points of similar phase, such as crest to crest.
By identifying and calculating these parameters, we can gain deeper insights into the specific behavior and properties of the wave in various contexts.
Wave Speed
Wave speed describes how quickly a wave travels through a medium and is dictated by the formula:\[ v = f \cdot \lambda \]This indicates that the speed \( v \) of a wave is the product of its frequency \( f \) and wavelength \( \lambda \). In our exercise, the calculated speed is 6.277 m/s.
Wave speed is relevant in various applications, from sound traveling through air to light in optical fibers. Understanding wave speed helps us predict how quickly information or energy is transmitted in different environments.Moreover, the medium's characteristics, like elasticity and density, significantly affect the speed. For instance, waves typically travel faster in solids compared to liquids, which in turn are faster than gases.
Power of a Wave
The power of a wave is the rate at which energy is transferred by the wave through a medium. It can be quantified using:\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]where:
  • \( \mu \) is the mass per unit length of the rope.
  • \( \omega \) is the angular frequency.
  • \( A \) is the amplitude.
  • \( v \) is the wave speed.
In practical terms, wave power is crucial in contexts such as harnessing energy from ocean waves or transmitting electrical signals.
In this exercise, the average power calculated is approximately 11.05 W.
Analyzing wave power allows us to design systems and applications that either maximize energy use or adequately handle the energy transmitted by waves. This concept is fundamental in fields like engineering and environmental science.

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Most popular questions from this chapter

A wire with mass 40.0 \(\mathrm{g}\) is stretched so that its ends are tied down at points 80.0 \(\mathrm{cm}\) apart. The wire vibrates in its fundamental mode with frequency 60.0 \(\mathrm{Hz}\) and with an amplitude at the antinodes of 0.300 \(\mathrm{cm} .\) (a) What is the speed of propagation oftransverse waves in the wire? (b) Compute the tension in the wire. (c) Find the maximum transverse velocity and acceleration of particles in the wire.

A piano wire with mass 3.00 \(\mathrm{g}\) and length 80.0 \(\mathrm{cm}\) is stretched with a tension of 25.0 \(\mathrm{N}\) . A wave with frequency 120.0 \(\mathrm{Hz}\) and amplitude 1.6 \(\mathrm{mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

15.40\(\cdot\) A 1.50 -m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 48.0 \(\mathrm{m} / \mathrm{s}\) . What are the wavelength and frequency of (a) the fundamental: (b) the second overtone; (c) the fourth harmonic?

CALC Equation \((15.7)\) for a sinusoidal wave can be made more general by including a phase angle \(\phi,\) where \(0 \leq \phi \leq 2 \pi(\) in radians). Then the wave function \(y(x, t)\) becomes $$y(x, t)=A \cos (k x-\omega t+\phi)$$ (a) Sketch the wave as a function of \(x\) at \(t=0\) for \(\phi=0,\) \(\phi=\pi / 4, \phi=\pi / 2, \phi=3 \pi / 4,\) and \(\phi=3 \pi / 2 .\) (b) Calculate the transverse velocity \(v_{y}=\partial y / \partial t .\) (c) At \(t=0,\) a particle on the string at \(x=0\) has displacement \(y=A / \sqrt{2} .\) Is this enough information to determine the value of \(\boldsymbol{\phi} ?\) In addition, if you are told that a particle at \(x=0\) is moving toward \(y=0\) at \(t=0,\) what is the value of \(\phi (\mathrm{d})\) Explain in general what you must know about the wave's behavior at a given instant to determine the value of \(\phi\) .

The upper end of a \(3.80-\) m-long steel wire is fastened to the ceiling, and a 54.0 -kg object is suspended from the lower end of the wire. You observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. What is the mass of the wire?

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