/*! 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 132 A traveling wave on a string is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A traveling wave on a string is modeled by the wave \(\quad\) equation \(y(x, t)=3.00 \mathrm{cm} \sin \left(8.00 \mathrm{m}^{-1} x+100.00 \mathrm{s}^{-1} t\right)\) string is under a tension of \(50.00 \mathrm{N}\) and has a linear mass density of \(\mu=0.008 \mathrm{kg} / \mathrm{m} .\) What is the average power transferred by the wave on the string?

Short Answer

Expert verified
The average power transferred by the wave on the string is approximately 9.00 Watts.

Step by step solution

01

Determine the wave speed

The wave speed can be found using the following formula: \(v = \sqrt{\frac{T}{\mu}}\), where v is the wave speed, T is the tension, and \(\mu\) is the linear mass density. Given the tension T = 50 N and the linear mass density \(\mu = 0.008 \, kg/m\), we can calculate the wave speed: \[v = \sqrt{\frac{50 N}{0.008 kg/m}}\]
02

Calculate the wave speed

Now let's calculate the wave speed: \[v = \sqrt{\frac{50 N}{0.008 kg/m}} = \sqrt{6250 m^2/s^2} = 25 m/s\]
03

Determine the amplitude and frequency of the wave

The wave equation is given by \(y(x,t) = 3.00 cm \sin(8.00 m^{-1} x + 100.00 s^{-1} t)\). The amplitude A is 3.00 cm, and the angular frequency \(\omega\) is 100 s^{-1}. To find the frequency f, we can use the relation \(\omega = 2\pi f\): \[f = \frac{\omega}{2\pi} = \frac{100 s^{-1}}{2\pi}\]
04

Calculate the frequency of the wave

Now let's calculate the frequency: \[f = \frac{100 s^{-1}}{2\pi} \approx 15.92 Hz\]
05

Calculate the average power transferred by the wave

Now that we have the amplitude, frequency, and wave speed of the wave, we can calculate the average power transferred by the wave using the formula: \(P_{avg} = \frac{1}{2}\mu v A^2 \omega^2\), where \(P_{avg}\) is the average power, \(\mu\) is the linear mass density, v is the wave speed, A is the amplitude, and \(\omega\) is the angular frequency. We have to convert the amplitude from centimeters to meters before plugging in the numbers: A = 3.00 cm = 0.03 m \[P_{avg} = \frac{1}{2}(0.008 kg/m)(25 m/s)(0.03 m)^2(100 s^{-1})^2\]
06

Find the average power transferred by the wave

Now let's find the average power transferred by the wave: \[P_{avg} = \frac{1}{2}(0.008 kg/m)(25 m/s)(0.03 m)^2(100 s^{-1})^2 \approx 9.00 W\] The average power transferred by the wave on the string is approximately 9.00 Watts.

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Key Concepts

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

Wave Speed
Wave speed is a crucial aspect of wave mechanics and helps define how fast a wave travels through a particular medium. In the context of waves on a string, the speed is determined by two main factors: the tension in the string and the linear mass density. The wave speed for a wave traveling along a string can be calculated using the formula:
  • Wave Speed, \(v = \sqrt{\frac{T}{\mu}}\)
Where:
  • \(v\) is the wave speed
  • \(T\) is the tension in the string
  • \(\mu\) is the linear mass density
By substituting the given values, T = 50 N and \(\mu = 0.008 \, kg/m\), into the formula, we find that the wave speed is \(25 \, m/s\). This result indicates that the wave travels 25 meters in one second along the string.
Average Power
The average power of a wave denotes the amount of energy that is transferred by the wave per unit time. Specifically, for mechanical waves like those traveling along a string, the power relates to how effectively the energy is being carried through the medium. The formula to determine the average power \(P_{avg}\) transferred by a wave on a string is:
  • \(P_{avg} = \frac{1}{2}\mu v A^2 \omega^2\)
Where:
  • \(\mu\) is the linear mass density
  • \(v\) is the wave speed
  • \(A\) is the amplitude of the wave
  • \(\omega\) is the angular frequency
To apply this formula correctly, convert the amplitude from centimeters to meters. Given the problem's data, we calculate the average power of the wave as approximately 9.00 Watts, which tells us how much energy is moving through the string per second.
Amplitude and Frequency
Amplitude and frequency are key characteristics of waves, each describing different aspects of a wave's motion. The amplitude of a wave signifies its maximum displacement from the rest position. In this case, the amplitude is given by the coefficient in the sine function of the wave equation, 3.00 cm, which we convert to meters as 0.03 m.Frequency describes how many cycles of a wave occur per second. The angular frequency \(\omega\) from the wave equation is related to the frequency \(f\) by:
  • \(f = \frac{\omega}{2\pi}\)
Given \(\omega = 100 \, s^{-1}\), this relation helps us find the frequency as approximately 15.92 Hz. Thus, the wave completes 15.92 cycles every second as it travels along the string.
Linear Mass Density
Linear mass density is a measure of mass per unit length for a string and has significant implications in determining the wave properties on that string. It is denoted by the symbol \(\mu\) and expressed in units of \(kg/m\). In wave mechanics, linear mass density is a key factor when calculating wave speed and power.Higher linear mass density means a greater mass for the same length of string, which tends to reduce wave speed because extra mass increases inertia, resulting in slower wave propagation.In the given problem, a linear mass density of \(0.008 \, kg/m\) was provided, and it plays a fundamental role in both calculating the wave speed and the power transmitted by the wave. Understanding linear mass density helps predict how variations in this property could impact wave behavior on strings.

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Most popular questions from this chapter

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of \(500.00 \mathrm{N}\) applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the \(3.00 \mathrm{m}\) of the string?

(a) Seismographs measure the arrival times of earthquakes with a precision of 0.100 s. To get the distance to the epicenter of the quake, geologists compare the arrival times of S-and P-waves, which travel at different speeds. If S- and P-waves travel at 4.00 and \(7.20 \mathrm{km} / \mathrm{s}\), respectively, in the region considered, how precisely can the distance to the source of the earthquake be determined? (b) Seismic waves from underground detonations of nuclear bombs can be used to locate the test site and detect violations of test bans. Discuss whether your answer to (a) implies a serious limit to such detection. (Note also that the uncertainty is greater if there is an uncertainty in the propagation speeds of the S- and P-waves.)

The speed of a transverse wave on a string is 300.00 \(\mathrm{m} / \mathrm{s},\) its wavelength is \(0.50 \mathrm{m},\) and the amplitude is 20.00 cm. How much time is required for a particle on the string to move through a distance of \(5.00 \mathrm{km}\) ?

A 2.40-m wire has a mass of 7.50 g and is under a tension of \(160 \mathrm{N}\). The wire is held rigidly at both ends and set into oscillation. (a) What is the speed of waves on the wire? The string is driven into resonance by a frequency that produces a standing wave with a wavelength equal to 1.20 m. (b) What is the frequency used to drive the string into resonance?

To increase the intensity of a wave by a factor of 50 by what factor should the amplitude be increased?

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