Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Transverse waves on a string have wave speed 8.00 m/s, amplitude 0.0700 m, and wavelength 0.320 m. The waves travel in the \(-x\)-direction, and at \(t = 0\) the \(x = 0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at \(x = 0.360\) m at time \(t = 0.150\) s. (d) How much time must elapse from the instant in part (c) until the particle at \(x = 0.360\) m next has maximum upward displacement?

Short Answer

Expert verified
Frequency: 25 Hz, period: 0.04 s, wave number: 19.63 m⁻¹. Next max at 0.190 s.

Step by step solution

01

Calculate Wave Frequency

The wave speed \(v\) is related to the wavelength \(\lambda\) and frequency \(f\) by the formula:\[ v = f \cdot \lambda \]Given the wave speed \(v = 8.00 \text{ m/s}\) and wavelength \(\lambda = 0.320 \text{ m}\), we can solve for the frequency:\[ f = \frac{v}{\lambda} = \frac{8.00}{0.320} = 25.0 \text{ Hz} \]
02

Calculate Wave Period

The period \(T\) of a wave is the inverse of the frequency:\[ T = \frac{1}{f} = \frac{1}{25.0} = 0.040 \text{ s} \]
03

Calculate Wave Number

The wave number \(k\) is given by the formula:\[ k = \frac{2\pi}{\lambda} \]Substituting the known wavelength:\[ k = \frac{2\pi}{0.320} \approx 19.63 \text{ m}^{-1} \]
04

Write the Wave Function

The general form of a wave traveling in the negative \(x\)-direction is:\[ y(x,t) = A \sin(kx + \omega t + \phi) \]where \(A = 0.0700 \text{ m}\), \(\omega = 2\pi f = 2\pi \times 25.0 = 50\pi \text{ rad/s}\), and the initial condition specifies maximum upward displacement at \(x = 0\) and \(t = 0\), so \(\phi = \frac{\pi}{2}\) (since sine of \(\pi/2\) gives maximum value). Thus the wave function becomes:\[ y(x,t) = 0.0700 \sin(19.63x + 50\pi t + \frac{\pi}{2}) \]
05

Find Transverse Displacement at Given Position and Time

Substitute \(x = 0.360 \text{ m}\) and \(t = 0.150 \text{ s}\) into the wave function:\[ y(0.360, 0.150) = 0.0700 \sin(19.63 \times 0.360 + 50\pi \times 0.150 + \frac{\pi}{2}) \]Calculating the argument:\[ 19.63 \times 0.360 + 50\pi \times 0.150 + \frac{\pi}{2} = 7.0676 + 23.5619 + 1.5708 \approx 32.2003 \text{ rad} \]\[ y = 0.0700 \sin(32.2003) \approx 0.026 \text{ m} \]
06

Calculate Time for Next Maximum Displacement

The time interval \(\Delta t\) between successive maximum displacements is the period \(T\), and the sine function reaches its maximum at one complete period later. Thus, from \(t = 0.150 \text{ s}\), the next maximum displacement occurs at:\[ t_{\text{next max}} = 0.150 + 0.040 = 0.190 \text{ s} \]Time elapsing from given time to next maximum:\[ \Delta t = 0.190 - 0.150 = 0.040 \text{ s} \]

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 Vaia!

Key Concepts

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

Transverse Waves
Transverse waves are fascinating waves that move perpendicular to the direction of wave propagation. Imagine holding a rope and moving your hand up and down; the waves travel horizontally while the medium (rope) moves vertically.
This is exactly how transverse waves behave.
  • An example of transverse waves can be seen in water waves or electromagnetic waves.
  • In the context of waves on a string, the particles of the string move up and down as the wave travels along the string.
  • These waves are different from longitudinal waves where the movement of the medium is parallel to wave travel, like sound waves in air.
In our exercise, these transverse waves on the string have a specific wave speed, amplitude, and they travel in the opposite of the positive x-direction, or (-x)-direction, showcasing the versatile nature of how waves can move across mediums.
Wave Speed
Wave speed is a crucial aspect of understanding how fast a wave travels through a medium. It determines how quickly the wave displaces particles along its path.
To find wave speed in a real-world scenario, it's vital to understand the relationship between wave speed, frequency, and wavelength:
  • The formula to find wave speed is: \[ v = f \cdot \lambda \]Where:
    • \( v \) is the wave speed.
    • \( f \) is the frequency, or the number of wave cycles per second.
    • \( \lambda \) is the wavelength, or the space between repeating units of wave pattern.
In the original problem, we found that the wave speed of the transverse wave on the string is 8.00 m/s. Knowing this makes it easier to calculate related properties like frequency and period using given physical quantities.
Wavelength
Wavelength is a measure of the distance between two identical phases of a wave, such as from crest to crest or trough to trough. It defines the length of a full wave cycle in a particular medium.
  • Physically, the wavelength can affect how waves interact with each other and with obstacles.
  • In our exercise, the given wavelength is 0.320 m, an important starting parameter. It helps us calculate other properties like the wave number.
  • The wavelength also impacts the wave speed and frequency, based on the formula \[ v = f \cdot \lambda \], showing that all these characteristics are intertwined.
By understanding the wavelength, we gain insight into how far a wave must travel before repeating itself, which is essential in wave physics applications.
Wave Function
A wave function is a mathematical description of the wave's oscillation in a medium as it travels. It provides a detailed explanation of displacement over time and space.
  • The general form of the wave function is: \[ y(x, t) = A \sin(kx + \omega t + \phi) \]where:
    • \( A \) is the amplitude, or maximum displacement.
    • \( k \) is the wave number, related to wavelength.
    • \( \omega \) is the angular frequency, found from \( \omega = 2\pi f \).
    • \( \phi \) is the phase constant, accounting for initial displacement.
  • For our problem, the wave function quantifies how the wave travels in the negative x-direction: \[ y(x, t) = 0.0700 \sin(19.63x + 50\pi t + \frac{\pi}{2}) \]
  • By using the wave function, we can predict the position of any particle on the string at any time.
Understanding wave functions is fundamental in wave physics since they allow us to model and anticipate wave behaviors naturally observed or in laboratory settings.

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

A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first overtone, a point at an antinode of the standing wave on the string has an amplitude of 3.50 cm and a maximum transverse speed of 28.0 m/s. (a) What is the string's mass? (b) What is the magnitude of the maximum transverse acceleration of this point on the string?

A simple harmonic oscillator at the point \(x = 0\) generates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function \(y(x, t)\) for the wave. Assume that the oscillator has its maximum upward displacement at time \(t = 0\). (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 Hz to about 20.0 kHz. (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short- wavelength sound, how far apart would the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meter stick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 m/s. How far apart would the dots be in each set? Could you readily measure their separation with a meter stick?

At a distance of \(7.00 \times 10^{12}\) m from a star, the intensity of the radiation from the star is 15.4 W/m\(^2\). Assuming that the star radiates uniformly in all directions, what is the total power output of the star?

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.

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.

Sign-up for free