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Chapter 17: Problem 42

Second Harmonic A rope, under a tension of \(200 \mathrm{~N}\) and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by $$ y(x, t)=(0.10 \mathrm{~m})\left(\sin \left(\left(\frac{\pi}{2} \mathrm{rad} / \mathrm{m}\right) x\right) \sin ((12 \pi \mathrm{rad} / \mathrm{s}) t)\right. $$ where \(x=0.00 \mathrm{~m}\) at one end of the rope, \(x\) is in meters, and \(t\) is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

Short Answer

Expert verified
a) 2 m, b) 24 m/s, c) 0.694 kg, d) 0.111 s

Step by step solution

01

Determine the wavelength

The given displacement equation for the standing wave pattern is y(x, t) = 0.10 \text{ m} \times \left(\sin(\left(\frac{\pi}{2} \text{rad}/\text{m}\right)x) \sin((12\pi \text{rad}/\text{s}) t)\right) \The term \(\sin(\left(\frac{\pi}{2} \text{rad}/\text{m}\right)x)\) indicates the spatial dependence of the wave. \Since this function represents the second harmonic, \frac{\pi}{2} \text{rad} / \text{m} = \frac{2\pi}{\lambda} we solve for \lambda: \[\frac{\pi}{2} = \frac{2\pi}{\lambda}\Rightarrow\lambda = 4 \text{ m}\]
02

Find the length of the rope

In the second harmonic, the wavelength is given by \(L = \frac{\lambda}{2}\), thus the length L of the rope becomes: \[L = \frac{4 \text{ m}}{2} = 2 \text{ m}\]
03

Calculate the speed of the wave

The speed v of the wave is given by the relationship \(v = f\lambda\). The wave frequency f is derived from the term \sin((12\pi \text{rad}/s) t), with the angular frequency \omega = 12\pi \text{rad}/s. \Therefore, \omega = 2 \pi f \Rightarrow f = \frac{12\pi}{2\pi} = 6 \text{Hz}\. Subsequently, the speed is: \[v = 6 \text{Hz} \times 4 \text{ m} = 24 \text{ m/s}\]
04

Find the mass of the rope

The mass m of the rope can be found using the tension T and the wave speed v in the formula: \[v = \sqrt{\frac{T}{\mu}} \Rightarrow \mu = \frac{T}{v^2} \] \where \mu = m/L\ is the linear mass density. Solving for \mu: \[\mu = \frac{200 \text{ N}}{(24 \text{ m/s})^2} = \frac{200}{576} \text{ kg/m} = 0.347 \text{ kg/m} \] \Finally, the mass m is: \[m = \mu \times L = 0.347 \text{ kg/m} \times 2 \text{ m} = 0.694 \text{ kg} \]
05

Determine the period for the third harmonic

For the third harmonic, the frequency f is given by: \[f_3 = 3 \times f_1 = 3 \times 3 \text{ Hz} = 9 \text{ Hz} \] \The period T is then: \[T = \frac{1}{f_3} = \frac{1}{9 \text{ Hz}} = \frac{1}{9} \text{ s} \approx 0.111 \text{ s} \]

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Key Concepts

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

Second-Harmonic Standing Wave
A second-harmonic standing wave refers to a vibration pattern in which the rope vibrates with two segments separated by one node (point of no motion). This pattern is characterized by having a wavelength \( \lambda \) that is twice the length of the rope \( L \). In mathematical terms, for the second harmonic: \[ L = \frac{\lambda}{2} \]. This means if the wavelength is 4 meters, the length of the rope is 2 meters.

The second-harmonic pattern results from the interference of two traveling waves moving in opposite directions with the same frequency and amplitude. These create a stable pattern of nodes and antinodes (points of maximum displacement). The displacement equation typically includes terms of sine functions for both position \( x \) and time \( t \). For instance, in our example, the equation is: \[ y(x, t)=(0.10 \, m) \left( \sin \left( \left( \frac{\pi}{2} \, rad / \, m \right) x \right) \sin (12 \pi \, rad / \, s \, t) \right) \].

This pattern of nodes and antinodes is called a standing wave because the pattern does not travel along the medium.
Wave Speed Calculation
Calculating the speed of a wave \( v \) involves understanding the wave's frequency \( f \) and its wavelength \( \lambda \). The simple formula to determine wave speed is: \[ v = f \times \lambda \].

First, find the angular frequency \( \omega \) from the given wave equation. Here, \( \omega = 12 \pi \, rad/s \). To get the frequency, use the relation \[ \omega = 2 \pi f \]. Rearrange it to find \[ f = \frac{\omega}{2 \pi} = \frac{12 \pi}{2 \pi} = 6 \, Hz \].

Next, multiply this frequency by the wavelength to get the wave speed: \[ v = 6 \, Hz \times 4 \, m = 24 \, m/s \].

This shows the relationship between how often the wave oscillates (frequency) and the physical distance between successive crests (wavelength) to determine its speed. This calculation is crucial for understanding how fast the energy or the wave pattern travels along the medium.
Mass and Tension Relationship
The speed of a wave on a rope or string is also related to the tension of the rope \( T \) and its linear mass density \( \mu \) (mass per unit length). This relationship can be described by the formula: \[ v = \sqrt{ \frac{T}{\mu} } \].

First, find the linear mass density by re-arranging the formula: \[ \mu = \frac{T}{v^2} \]. With a given tension of \( T = 200 \, N \) and wave speed of \( v = 24 \, m/s \), calculate \[ \mu = \frac{200 \, N}{(24 \, m/s)^2} = \frac{200}{576} \, kg/m = 0.347 \, kg/m \].

Finally, the total mass \( m \) of the rope is obtained by multiplying the linear mass density by the length \( L \) of the rope: \[ m = \mu \times L = 0.347 \, kg/m \times 2 \, m = 0.694 \, kg \].

This demonstrates how the characteristics of the wave and the tension applied to the medium collectively determine the mass distribution along the rope. It also emphasizes the importance of these physical properties in wave mechanics.

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Most popular questions from this chapter

Show That Show that $$ \begin{array}{ll} y(x, t)=Y \sin k(x-v t), & y(x, t)=Y \sin 2 \pi\left(\frac{x}{\lambda}-f t\right) \\ y(x, t)=Y \sin \omega\left(\frac{x}{v}-t\right), & y(x, t)=Y \sin 2 \pi\left(\frac{x}{\lambda}-\frac{t}{T}\right) \end{array} $$ are all equivalent to \(y(x, t)=Y \sin (k x-\omega t)\).

Linear Density The linear density of a string is \(1.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). A transverse wave on the string is described by the equation $$ y(x, t)=(0.021 \mathrm{~m}) \sin [(2.0 \mathrm{rad} / \mathrm{m}) x+(30 \mathrm{rad} / \mathrm{s}) t] $$ What is (a) the wave speed and (b) the tension in the string?

The Equation of The equation of a transverse wave on a string is $$ y(x, t)=(2.0 \mathrm{~mm}) \sin [(20 \mathrm{rad} / \mathrm{m}) x-(600 \mathrm{rad} / \mathrm{s}) t] $$ The tension in the string is \(15 \mathrm{~N}\). (a) What is the wave speed? (b) Find the linear density of the string in grams per meter.

Equation of a Transverse The equation of a transverse wave traveling along a very long string is \(y(x, t)=(6.0 \mathrm{~cm}) \sin \\{(0.020 \mathrm{\pi} \mathrm{rad} / \mathrm{cm})\) \(x+(4.0 \pi \mathrm{rad} / \mathrm{s}) t]\) where \(x\) and \(y\) are expressed in centimeters and \(t\) is in seconds Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at \(x=3.5 \mathrm{~cm}\) when \(t=0.26 \mathrm{~s}\) ?

Aluminum Wire In Fig. 17-38, an aluminum wire, of length \(L_{1}=60.0\) \(\mathrm{cm}\), cross-sectional area \(1.00 \times 10^{-2} \mathrm{~cm}^{2}\), and den- sity \(2.60 \mathrm{~g} / \mathrm{cm}^{3}\), is joined to a steel wire, of density \(7.80 \mathrm{~g} / \mathrm{cm}^{3}\) and the same cross-sectional area. The compound wire, loaded with a block of mass \(m=10.0 \mathrm{~kg}\), is arranged so that the distance \(L_{2}\) from the joint to the supporting pulley is \(86.6 \mathrm{~cm} .\) Transverse waves are set up in the wire by using an external source of variable frequency; a node is located at the pulley. (a) Find the lowest frequency of excitation for which standing waves are observed such that the joint in the wire is one of the nodes. (b) How many nodes are observed at this frequency?
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