Skip to main content
Ch. 15 - Wave Motion
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 15 - Wave MotionProblema 55a
Capitolo 15, Problema 55a

The displacement of a standing wave on a string is given by D = 2.4 sin ( 0.60x ) cos (42t) , where x and D are in centimeters and t is in seconds. What is the distance (cm) between nodes?

Guida verificata passo dopo passo
1
The equation of the standing wave is given as \( D = 2.4 \sin(0.60x) \cos(42t) \). In this equation, \( \sin(0.60x) \) represents the spatial part of the wave, and \( \cos(42t) \) represents the time-dependent part. Nodes occur where the displacement \( D \) is always zero, regardless of time.
To find the nodes, focus on the spatial part of the wave, \( \sin(0.60x) \). Nodes occur when \( \sin(0.60x) = 0 \). The sine function equals zero at integer multiples of \( \pi \), so \( 0.60x = n\pi \), where \( n \) is an integer (\( n = 0, \pm 1, \pm 2, \dots \)).
Solve for \( x \) to find the positions of the nodes: \( x = \frac{n\pi}{0.60} \). This gives the positions of the nodes along the string.
The distance between two consecutive nodes is the wavelength divided by 2. The wavelength \( \lambda \) is related to the wave number \( k \) by \( k = \frac{2\pi}{\lambda} \). From the equation \( \sin(0.60x) \), the wave number \( k \) is \( 0.60 \), so \( \lambda = \frac{2\pi}{0.60} \).
Finally, calculate the distance between nodes as \( \frac{\lambda}{2} = \frac{1}{2} \cdot \frac{2\pi}{0.60} \). This gives the distance between consecutive nodes in centimeters.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
3m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Standing Waves

Standing waves are formed when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other. This results in a wave pattern that appears to be stationary, characterized by fixed points called nodes (points of no displacement) and antinodes (points of maximum displacement). Understanding standing waves is crucial for analyzing wave behavior in strings and other mediums.
Video consigliato:
Percorso guidato
07:58
Intro to Transverse Standing Waves

Nodes and Antinodes

Nodes are specific points along a standing wave where the displacement is always zero, while antinodes are points where the displacement reaches its maximum. The distance between two consecutive nodes is half the wavelength of the wave. Identifying these points is essential for solving problems related to wave patterns and their properties.
Video consigliato:
Percorso guidato
08:24
Example 1

Wavelength

Wavelength is the distance between two consecutive points that are in phase on a wave, such as from one node to the next node. In the context of standing waves, the wavelength can be determined from the wave equation. For the given equation, the spatial component (0.60x) indicates the wave number, which can be used to calculate the wavelength and subsequently the distance between nodes.
Video consigliato:
Percorso guidato
05:42
Unknown Wavelength of Laser through Double Slit
Pratica correlata
Domanda del libro di testo

A guitar string is 91 cm long and has a mass of 3.2 g. The vibrating portion of the string from the bridge to the support post is ℓ = 64cm and the string is under a tension of 520 N. What are the frequencies of the fundamental and first two overtones?

1195
views
Domanda del libro di testo

The displacement of a standing wave on a string is given by D = 2.4sin(0.60x)cos(42t), where x and D are in centimeters and t is in seconds. Give the amplitude, frequency, and speed of each of the component waves.

975
views
Domanda del libro di testo

One end of a horizontal string is attached to a small-amplitude mechanical 60.0-Hz oscillator. The string’s mass per unit length is 3.9 x 10⁻ ⁴ kg/m. The string passes over a pulley, a distance ℓ = 1.50 m away, and weights are hung from this end, Fig. 15–38. What mass m must be hung from this end of the string to produce five loops of a standing wave? Assume the string at the oscillator is a node, which is nearly true.

1252
views
Domanda del libro di testo

A particular violin string plays at a frequency of 294 Hz. If the tension is increased 22%, what will the new frequency be?

1141
views
Domanda del libro di testo

The speed of waves on a string is 96 m/s. If the frequency of standing waves is 435 Hz, how far apart are two adjacent nodes?

1262
views