/*! 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 41 A thin, taut string tied at both... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t) = (5.60 \, \mathrm{cm}) \mathrm{sin} [(0.0340 \mathrm{rad/cm})x] \mathrm{sin} [(150.0 \, \mathrm{rad/s})t]\), where the origin is at the left end of the string, the x-axis is along the string, and the \(y\)-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x,t)\) for this string if it were vibrating in its eighth harmonic?

Short Answer

Expert verified
(a) Sketch nodes at x = 0, L/3, 2L/3, L. (b) 2.80 cm. (c) 276.09 cm. (d) \( \lambda = 184.06 \, \mathrm{cm}, f = 23.87 \, \mathrm{Hz}, v = 439.40 \, \mathrm{cm/s} \). (e) 840 cm/s. (f) \( y(x,t) = 5.60 \sin[(0.091)x]\sin[150.0t] \).

Step by step solution

01

Understanding the Standing Wave

A standing wave is typically formed by the superposition of two traveling waves moving in opposite directions. The given equation \( y(x, t) = (5.60 \, \mathrm{cm}) \sin[(0.0340 \, \mathrm{rad/cm})x] \sin[(150.0 \, \mathrm{rad/s})t] \) represents a standing wave.
02

Sketching the Standing Wave Pattern

The given equation describes the third harmonic, which has nodes at both ends and two additional nodes in between. Sketch the wave by marking nodes at points \( x = 0, L/3, 2L/3, L \) and antinodes at points \( x = L/6, L/2, 5L/6 \).
03

Amplitude of Traveling Waves

The amplitude of the standing wave is 5.60 cm. For a standing wave formed by two identical traveling waves moving in opposite directions, the amplitude of each traveling wave is half of the standing wave's amplitude. Thus, each traveling wave has an amplitude of \( 5.60 \, \mathrm{cm} / 2 = 2.80 \, \mathrm{cm} \).
04

Determining the Length of the String

In the third harmonic, the length of the string is \( L = \frac{3}{2} \lambda \). From the wave equation, the wave number \( k = 0.0340 \, \mathrm{rad/cm} \), so the wavelength \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.0340} \approx 184.06 \, \mathrm{cm} \). The length \( L = \frac{3}{2} \times 184.06 \approx 276.09 \, \mathrm{cm} \).
05

Calculating Wavelength, Frequency, and Speed

Wavelength \( \lambda = \frac{2\pi}{0.0340} = 184.06 \, \mathrm{cm} \). The angular frequency \( \omega = 150.0 \, \mathrm{rad/s} \) gives the frequency \( f = \frac{\omega}{2\pi} = \frac{150.0}{2\pi} \approx 23.87 \, \mathrm{Hz} \). The speed \( v = f \lambda = 23.87 \times 1.8406 \approx 439.40 \, \mathrm{cm/s} \).
06

Maximum Transverse Speed

The maximum transverse speed of a point on the string is given by the product of the amplitude and the angular frequency: \( v_{max} = A \omega = 5.60 \, \mathrm{cm} \times 150.0 \, \mathrm{rad/s} = 840 \, \mathrm{cm/s} \).
07

Equation for the Eighth Harmonic

For the eighth harmonic, the relationship between length and wavelength is \( L = 4\lambda \). So, \( \lambda = \frac{L}{4} = \frac{276.09}{4} = 69.0225 \, \mathrm{cm} \). The new wave number \( k = \frac{2\pi}{\lambda} \approx 0.091 \, \mathrm{rad/cm} \). The wave equation becomes \( y(x, t) = (5.60 \, \mathrm{cm}) \sin[(0.091 \, \mathrm{rad/cm})x] \sin[(150.0 \, \mathrm{rad/s})t] \).

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Key Concepts

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

Harmonics
In the world of waves, harmonics are special natural patterns that occur on a string fixed at both ends. When a string is vibrated, it doesn't just generate random waves; instead, specific patterns called harmonics are formed. The harmonics indicate the different ways the string can vibrate. Each harmonic corresponds to a standing wave, with the first harmonic being the fundamental frequency. The third harmonic, like in our exercise, means that the string oscillates with three segments. There are nodes at each end and additional nodes between these segments, where the string doesn't move, while the sections in between these nodes, called antinodes, experience the maximum displacement. Understanding harmonics helps us determine the characteristic frequencies at which a system can naturally oscillate.
Wave Equation
The wave equation is a mathematical way to describe how the shape of a wave changes over time and space. It's an essential part of understanding wave behavior. In general, a wave equation can be expressed in terms of a sine function due to its repetitive nature, showing how the wave repeats in cycles.
The equation given in the exercise, \(y(x, t) = (5.60 \, \mathrm{cm}) \sin[(0.0340 \, \mathrm{rad/cm})x] \sin[(150.0 \, \mathrm{rad/s})t]\) is specific to the standing wave on the string in its third harmonic. Here, the two sine functions represent the spatial and temporal components that describe the wave's shape as it oscillates over time. The wave number \(k = 0.0340 \, \mathrm{rad/cm}\) provides information about the wavelength, and the angular frequency \(\omega = 150.0 \, \mathrm{rad/s}\) relates to how fast the wave oscillates. Together, these parameters give a complete picture of the wave's behavior.
Transverse Speed
In waves on a string, the transverse speed is about how fast a point on the string moves up and down (perpendicular to the direction of the wave). Think of a surfer riding waves; the up and down motion is similar to the transverse speed. For any point on the string, the transverse speed can change as the wave passes through it. The maximum transverse speed is the highest speed at which any point moves vertically.
To calculate the maximum transverse speed, we use the formula \(v_{\text{max}} = A \omega\), where \(A\) is the amplitude of the wave and \(\omega\) is the angular frequency. From the exercise's wave equation, the amplitude \(A\) is \(5.60 \, \mathrm{cm}\), and \(\omega\) is \(150.0 \, \mathrm{rad/s}\). Hence, the maximum transverse speed \(v_{\text{max}}\) comes out to be \(840 \, \mathrm{cm/s}\). This gives us an understanding of the dynamic movement of the string as the waves travel along it.

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Most popular questions from this chapter

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t) = (8.50 \, \mathrm{mm}) \mathrm{cos}(172 \, \mathrm{rad/m} \space x - 4830 \, \mathrm{rad/s} \space t)$$ Assume that the tension of the string is constant and equal to \(W\). (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W\)? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling \(down\) the string?

One string of a certain musical instrument is 75.0 cm long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 m/s. (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 m? (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120 Hz. The other end passes over a pulley and supports a 1.50-kg mass. The linear mass density of the rope is 0.0480 kg/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 kg?

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal, stretched string with a speed of 36.0 m/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 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 transverse velocity of a particle 1.35 m to the right of the origin at time \(t = 0.0625\) s.

A certain transverse wave is described by $$y(x, t) = (6.50 \, \mathrm{mm}) \mathrm{cos} \, 2\pi \Big( \frac{x}{28.0 \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \Big)$$ Determine the wave's (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

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