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Chapter 15: Problem 24

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is \(y(x, t)=\) 2.30 \(\mathrm{mm} \cos [(6.98 \mathrm{rad} / \mathrm{m}) x+(742 \mathrm{rad} / \mathrm{s}) t] .\) Being more practical, you measure the rope to have a length of 1.35 \(\mathrm{m}\) and a mass of 0.00338 \(\mathrm{kg}\) . You are then asked to determine the following: (a) amphtude; (b) frequency; (c) wavelength; (d) wave speed; (e) direction the wave is traveling; (f) tension in the rope; (g) aver- age power transmitted by the wave.

Short Answer

Expert verified
Amplitude: 2.30 mm; Frequency: ~118.16 Hz; Wavelength: ~0.90 m; Wave Speed: ~106.34 m/s; Direction: Negative x-direction; Tension: ~28.3 N; Power: ~5.04 W.

Step by step solution

01

Identify the Amplitude

The amplitude of the wave can be directly read from the wave function given. The term outside the cosine function represents the amplitude. In this case, the amplitude \( A \) is 2.30 mm.
02

Determine the Frequency

The angular frequency \( \omega \) is given as 742 rad/s. The frequency \( f \) is related to the angular frequency by the formula \( f = \frac{\omega}{2\pi} \). Substituting the known value, \( f = \frac{742}{2\pi} \approx 118.16 \) Hz.
03

Calculate the Wavelength

The wave number \( k \) is given as 6.98 rad/m. The wavelength \( \lambda \) is related to the wave number by the formula \( \lambda = \frac{2\pi}{k} \). Substituting the known value, \( \lambda = \frac{2\pi}{6.98} \approx 0.90 \) m.
04

Find the Wave Speed

The wave speed \( v \) is related to the frequency and the wavelength by the equation \( v = f \cdot \lambda \). Substituting the known values, \( v = 118.16 \text{ Hz} \times 0.90 \text{ m} \approx 106.34 \text{ m/s} \).
05

Determine the Direction of the Wave

The wave function contains the term \((6.98 \, \text{rad/m})x + (742 \, \text{rad/s})t\). Since both the spatial and temporal components have the same sign, the wave travels in the negative x-direction.
06

Calculate the Tension in the Rope

To find the tension, we first calculate the linear mass density \( \mu = \frac{\text{mass}}{\text{length}} = \frac{0.00338 \, \text{kg}}{1.35 \, \text{m}} \approx 0.0025 \, \text{kg/m}\). The tension \( T \) can be found using the wave speed formula \( v = \sqrt{\frac{T}{\mu}} \). Solving for \( T \), we get \( T = v^2 \cdot \mu \approx 106.34^2 \cdot 0.0025 \approx 28.3 \, \text{N} \).
07

Calculate the Average Power Transmitted by the Wave

The power is calculated using the formula \( P = \frac{1}{2} \mu v \omega^2 A^2 \), where \( A \) is the amplitude in meters. Substituting the values, \( P = \frac{1}{2} \times 0.0025 \times 106.34 \times 742^2 \times (2.30 \times 10^{-3})^2 \approx 5.04 \,\text{W}\).

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Key Concepts

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

Frequency Calculation
Understanding frequency is essential in wave mechanics. In the provided exercise, the angular frequency \( \omega \) is given as 742 rad/s. Frequency, denoted by \( f \), can be understood as the number of oscillations or cycles that occur per second. To find \( f \), we use the relation \( f = \frac{\omega}{2\pi} \). Here, \( \omega = 742 \) rad/s, hence:
  • \( f = \frac{742}{2\pi} \approx 118.16 \, \text{Hz} \)
This value indicates how many wave cycles pass a fixed point per second.
Wavelength Determination
Wavelength is the distance between consecutive points of a wave, such as crest to crest or trough to trough. In this exercise, the wave number \( k \) is given as 6.98 rad/m, which helps calculate the wavelength \( \lambda \). The relationship is \( \lambda = \frac{2\pi}{k} \).
  • \( \lambda = \frac{2\pi}{6.98} \approx 0.90 \, \text{m} \)
This means each wave is 0.90 meters long, a fundamental property of the wave's spatial extent.
Wave Speed Calculation
Wave speed tells us how fast a wave travels through a medium. It is determined by the product of frequency and wavelength: \( v = f \cdot \lambda \). With frequency \( f = 118.16 \) Hz and wavelength \( \lambda = 0.90 \) m:
  • \( v = 118.16 \, \text{Hz} \times 0.90 \, \text{m} \approx 106.34 \, \text{m/s} \)
This calculation shows the speed at which the wave propagates along the rope.
Mechanical Waves
Mechanical waves require a medium to travel. In this exercise, the wave travels on a rope, making it a classic example of a mechanical wave. These waves can be transverse or longitudinal depending on how they displace the medium. The given wave function indicates a transverse wave, as the displacement is perpendicular to the direction of wave propagation.
Mechanical waves showcase interesting properties such as reflection, refraction, and interference as they move through various media. Understanding these helps in fields ranging from acoustics to earthquake studies.
Tension Calculation
Calculating the tension in the rope involves understanding the relationship between wave speed and tension. From our calculations, the linear mass density \( \mu \) is \( \frac{0.00338}{1.35} \approx 0.0025 \, \text{kg/m} \). The tension \( T \) can be derived from the equation \( v = \sqrt{\frac{T}{\mu}} \).
  • \( T = v^2 \cdot \mu \approx 106.34^2 \cdot 0.0025 \approx 28.3 \, \text{N} \)
This provides insight into the force exerted along the rope, which is crucial for understanding wave propagation dynamics.
Power Transmission in Waves
The power transmitted by a wave indicates how much energy is being transferred by the wave per second. For this wave on a rope, power \( P \) is given by the formula:
  • \( P = \frac{1}{2} \mu v \omega^2 A^2 \)
With \( \mu = 0.0025 \text{ kg/m} \), \( v = 106.34 \text{ m/s} \), \( \omega = 742 \text{ rad/s} \), and amplitude \( A = 2.30 \times 10^{-3} \text{ m} \):
  • \( P \approx 5.04 \, \text{W} \)
This illustrates the capacity of the wave to do work or transmit energy across the medium, vital for applications like communication and power transmission systems.

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Most popular questions from this chapter

Wave Equation and Standing Waves. (a) Prove by direct substitution that \(y(x, t)=\left(A_{\text { sw }} \sin k x\right)\) sin \(\omega t\) is a solution of the wave equation, Eq. \((15.12),\) for \(v=\omega / k .\) (b) Explain why the relationship \(v=\omega / k\) for traveling waves also applies to standing waves.

Waves on a Stick. A flexible stick 2.0 \(\mathrm{m}\) long is not fixed in any way and is free to vibrate. Make clear drawings of this stick vibrating in its first three harmonics, and then use your drawings to find the wavelengths of each of these harmonics. (Hint: Should the ends be nodes or antinodes?)

Tsunami! On December \(26,2004,\) a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some \(200,000\) people. Satellites observing these waves from space measured 800 \(\mathrm{kn}\) from one wave crest to the next and a period between waves of 1.0 hour. What was the speed of these waves in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\) ? Does your answer help you understand why the waves caused such devastation?

Holding Up Under Stress. A string or rope will braak apart if it is placed under too much tensile stress \([\mathrm{Eq} \text { . }(11.8)]\) . Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density 7800 \(\mathrm{kg} / \mathrm{m}^{3}\) and will break if the tensile stress exceeds \(7.0 \times 10^{8} \mathrm{N} / \mathrm{m}^{2} .\) You want to make a guitar string from 4.0 \(\mathrm{g}\) of this type of steel. In use, the guitar string must be able to withstand a tension of 900 \(\mathrm{N}\) without breaking. Your job is the following: (a) Determine the maximum length and minimum radius the string can have. (b) Determine the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.

Energy in a Triangular Pulse. A triangular wave pulse on a taut string travels in the positive \(x\) -direction with speed \(v\) . The tension in the string is \(F,\) and the linear mass density of the string is \(\mu .\) At \(t=0,\) the shape of the pulse is given by $$ y(x, 0)=\left\\{\begin{array}{ll}{0} & {\text { if } x<-L} \\ {h(L+x) / L} & {\text { for }-LL}\end{array}\right. $$ (a) Draw the pulse at \(t=0 .\) (b) Determine the wave function \(y(x, t)\) at all times \(t\) (c) Find the instantancous power in the wave. Show that the power is zero except for \(-L<(x-v t)
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