/*! 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 39 A thin, taut string tied at both... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t)=(5.60 \mathrm{~cm}) \sin [(0.0340 \mathrm{rad} / \mathrm{cm}) x] \sin [(50.0 \mathrm{rad} / \mathrm{s}) t],\) 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) The standing wave pattern would show three complete wavelengths. b) The amplitude of the two travelling waves is 5.60 cm. c) The length of the string is 277 cm. d) The wavelength is 585 cm, the frequency is 8 Hz, the period is 0.125 s, and the speed is 4680 cm/s. e) The maximum transverse speed of a point on the string is 280 cm/s. f) The equation for the string vibrating in its eight harmonic is \(y_8(x, t) = 5.60 cm \sin[8(0.0340 rad/cm)x] \sin[50.0 rad/s t]\).

Step by step solution

01

Identify Known Values

The given equation for the wave is \(y(x, t) = (5.60 cm) \sin[(0.0340 rad/cm)x] \sin[(50.0 rad/s)t]\). From this equation and the problem, it can be noted that the amplitude (A) is 5.60 cm, the angular wave number (\(k\)) is 0.0340 rad/cm, and the angular frequency (\(ω\)) is 50.0 rad/s.
02

Calculate Amplitude

For a standing wave, the amplitude is the same as the amplitude of the two identical traveling waves that make it up. Therefore, the amplitude of the traveling waves is also 5.60 cm.
03

Find Length of the String

In the third harmonic, the string length equals three half-wavelengths. Therefore, \(L=3λ/2\). However, since \(k = 2π/λ\), we find that \(λ = 2π/k\). Substituting this into the above equation gives \(L = 3*(2π/k)/2 = 3π/k\). Substituting the known value of k, we find that \(L = 3π/0.0340 rad/cm = 277 cm\).
04

Find Wavelength, Frequency, Period, and Speed

From before, we know that λ = 2π/k. Therefore, λ = 2π/0.0340 rad/cm = 585 cm. To find the frequency, note that ω = 2πν, and therefore, ν = ω/2π = 50.0 rad/s / 2π = 8 Hz. The period T is the reciprocal of the frequency, so T = 1/ν = 0.125 s. The wave speed v can be found from the relationship v = λν, giving v = (585 cm)(8 Hz) = 4680 cm/s.
05

Calculate Maximum Transverse Speed

The speed of a point on the string is v = ωA. Substituting the known values of ω and A, we find that v = (5.60 cm)(50.0 rad/s) = 280 cm/s.
06

Eighth Harmonic Equation

For the eight harmonic, y(x, t) is changed by replacing the wave number with eight times its value. Therefore, for the eight harmonic, \(y_8(x, t) = 5.60 cm \sin[8(0.0340 rad/cm)x] \sin[50.0 rad/s t]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Standing Waves
Standing waves are a fascinating phenomenon that occurs when waves traveling in opposite directions interfere with each other in a fixed medium. In the context of a string tied at both ends, like a guitar string, standing waves form due to the reflection of waves at the rigid boundaries.
This results in specific points, called nodes, where there is no motion, and other points, called antinodes, where the motion is maximum.
  • The wave 'stands still' rather than traveling, hence the name 'standing wave'.
  • These waves can be visualized as patterns formed on the string when it vibrates in one of its harmonics.
In the given exercise, the string is oscillating in its third harmonic. This means there are three antinodes and nodes are positioned in between along the length of the string.
Standing wave patterns are essential in understanding how musical instruments produce sound, providing a natural resonance that amplifies the sound.
Harmonics
Harmonics are essentially the natural frequencies at which a system resonates. For a string fixed at both ends, these harmonics relate to the integer multiples of its fundamental frequency.
  • The fundamental frequency (first harmonic) is the lowest frequency of vibration.
  • Higher harmonics correspond to higher frequencies that are integer multiples of the fundamental.
In the exercise, the string is oscillating in its third harmonic, meaning it is vibrating at three times the fundamental frequency.
Each harmonic has its characteristic pattern with a different number of nodes and antinodes. For instance, the third harmonic has two additional nodes compared to the fundamental frequency.
Understanding harmonics is crucial for designing musical instruments, as they determine the pitch and timbre of the sound produced.
Wave Equation
The wave equation provided in the exercise, \(y(x, t) = (5.60 \text{ cm}) \sin[(0.0340 \text{ rad/cm}) x] \sin[(50.0 \text{ rad/s}) t]\), describes the displacement of the string at any point \(x\) and time \(t\). This equation is typical for a standing wave formed from two traveling waves of identical frequency and amplitude moving in opposite directions.
  • It consists of two sine functions: one dependent on space \(x\) and the other on time \(t\).
  • The amplitude \(A = 5.60 \text{ cm}\) indicates the maximum displacement from the rest position.
Breaking down the equation:
  • The term \(0.0340 \text{ rad/cm}\) represents the wave number \(k\), which is related to the wavelength by \(k = \frac{2\pi}{\lambda}\).
  • The term \(50.0 \text{ rad/s}\) represents the angular frequency \(\omega\), related to the frequency \(u\) by \(\omega = 2\piu\).
Solving this equation helps us understand the characteristics of the wave, such as wavelength, frequency, and speed.
Transverse Waves
Transverse waves are a type of wave where the displacement of the medium is perpendicular to the direction of wave propagation. In the context of the exercise, the wave on the string is a transverse wave because the string moves up and down while the wave travels along its length.
This type of wave is common in vibrating strings and electromagnetic waves.
  • In a transverse wave on a string, peaks and troughs are created as the wave travels, showing motion perpendicular to the string's length.
  • Unlike longitudinal waves, where the displacement occurs in the direction of travel (such as sound waves in air), transverse waves are easily visible, making them simpler to analyze.
Understanding transverse waves is vital in many fields of physics, including optics and acoustics, where the nature of wave propagation dictates how energy and information are transferred through a medium.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Adjacent antinodes of a standing wave on a string are \(15.0 \mathrm{~cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude \(0.850 \mathrm{~cm}\) and period \(0.0750 \mathrm{~s}\). The string lies along the \(+x\) -axis and is fixed at \(x=0 .\) (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

A water wave traveling in a straight line on a lake is described by the equation $$ y(x, t)=(2.75 \mathrm{~cm}) \cos (0.410 \mathrm{rad} / \mathrm{cm} x+6.20 \mathrm{rad} / \mathrm{s} t) $$ where \(y\) is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? (b) What are the wave number and the number of waves per second that pass the fisherman? (c) How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?

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?

A \(1.005 \mathrm{~m}\) chain consists of small spherical beads, each with a mass of \(1.00 \mathrm{~g}\) and a diameter of \(5.00 \mathrm{~mm},\) threaded on an elastic strand with negligible mass such that adjacent beads are separated by a center-to-center distance of \(10.0 \mathrm{~mm}\). There are beads at each end of the chain. The strand has a spring constant of \(28.8 \mathrm{~N} / \mathrm{m}\). The chain is stretched horizontally on a frictionless tabletop to a length of \(1.50 \mathrm{~m}\), and the beads at both ends are fixed in place. (a) What is the linear mass density of the chain? (b) What is the tension in the chain? (c) With what speed would a pulse travel down the chain? (d) The chain is set vibrating and exhibits a standing-wave pattern with four antinodes. What is the frequency of this motion? (e) If the beads are numbered sequentially from 1 to \(101,\) what are the numbers of the five beads that remain motionless? (f) The 13th bead has a maximum speed of \(7.54 \mathrm{~m} / \mathrm{s}\). What is the amplitude of that bead's motion? (g) If \(x_{0}=0\) corresponds to the center of the 1 st bead and \(x_{101}=1.50 \mathrm{~m}\) corresponds to the center of the 101 st bead, what is the position \(x_{n}\) of the \(n\) th bead? (h) What is the maximum speed of the 30 th bead?

You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A \(2.00-\mathrm{cm}\) -long piece of the wire is left over from its installation. Using an analytical balance, you determine that the mass of the spare piece is \(14.5 \mu \mathrm{g}\). You then hang a \(0.400 \mathrm{~kg}\) mass from the lower end of the long, suspended wire. When a small-amplitude transverse wave pulse is sent up that wire, sensors at both ends measure that it takes the wave pulse \(26.7 \mathrm{~ms}\) to travel the length of the wire. (a) Use these measurements to calculate the length of the wire. Assume that the weight of the wire has a negligible effect on the speed of the transverse waves. (b) Discuss the accuracy of the approximation made in part (a).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.