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

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.

Short Answer

Expert verified
Frequency: 20 Hz; Angular Frequency: 40\(\pi\) rad/s; Wave Number: \(\frac{10\pi}{9}\) m\(^{-1}\).

Step by step solution

01

Calculate the Frequency

To find the frequency, we use the wave speed formula: \( v = f \lambda \). Here, \( v = 36.0 \) m/s and \( \lambda = 1.80 \) m. Solving for \( f \), we get \( f = \frac{v}{\lambda} = \frac{36.0}{1.80} = 20.0 \) Hz.
02

Determine the Angular Frequency

The angular frequency, \( \omega \), is related to the frequency by \( \omega = 2\pi f \). Substituting \( f = 20.0 \) Hz, we get \( \omega = 2\pi \times 20.0 = 40\pi \) rad/s.
03

Find the Wave Number

The wave number, \( k \), is given by \( k = \frac{2\pi}{\lambda} \). Substituting \( \lambda = 1.80 \) m, we calculate \( k = \frac{2\pi}{1.80} = \frac{10\pi}{9} \) m\(^{-1}\).
04

Write the Wave Function

The wave function is given by \( y(x, t) = A \sin(kx - \omega t + \phi) \). Since at \( t = 0 \), \( x = 0 \), and the displacement is maximum, \( \phi = \frac{\pi}{2} \). Thus, \( y(x, t) = 2.5 \sin(\frac{10\pi}{9}x - 40\pi t + \frac{\pi}{2}) \).
05

Find the Function at the Left End

For a particle at the left end, \( x = 0 \). Thus, the function simplifies to \( y(0, t) = 2.5 \sin(-40\pi t + \frac{\pi}{2}) \).
06

Function for Particle 1.35 m to the Right

For a particle 1.35 m to the right, \( x = 1.35 \) m. Substitute into the wave function: \( y(1.35, t) = 2.5 \sin(\frac{10\pi}{9} \times 1.35 - 40\pi t + \frac{\pi}{2}) \).
07

Maximum Transverse Velocity

The maximum transverse velocity, \( v_{max} \), is given by \( A \omega \). Using \( A = 2.5 \) mm = 0.0025 m and \( \omega = 40\pi \) rad/s, we get \( v_{max} = 0.0025 \times 40\pi = 0.1\pi \) m/s.
08

Displacement and Velocity at Specific Time

At \( t = 0.0625 \) s and \( x = 1.35 \) m, substitute into the wave function to find displacement: \( y(1.35, 0.0625) = 2.5 \sin(\frac{10\pi}{9} \times 1.35 - 40\pi \times 0.0625 + \frac{\pi}{2}) \). To find the velocity, differentiate \( y(x, t) \) with respect to \( t \) and substitute \( t = 0.0625 \), \( x = 1.35 \).

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.

Wave Frequency
Wave frequency is a fundamental aspect of wave physics, representing how many waves pass a point per second. In our problem, we calculated the frequency using the formula:
  • Formula: \( v = f \lambda \)
  • Where \( v \) is wave speed, \( f \) is frequency, and \( \lambda \) is wavelength.
Inserting the given values, \( v = 36.0 \text{ m/s} \) and \( \lambda = 1.80 \text{ m} \), we solve for frequency:\[ f = \frac{36.0}{1.80} = 20.0 \text{ Hz} \]This outcome tells us that 20 waves pass by any point on the string every second. Understanding wave frequency helps in deciphering how fast a wave is oscillating, which is crucial in predicting wave behaviors in different environments.
Angular Frequency
Angular frequency, denoted as \( \omega \), relates to how quickly the wave oscillates in terms of radians per second. Unlike the normal frequency, it uses radians to describe oscillations rather than cycles.
  • Formula: \( \omega = 2\pi f \)
  • \( f \) is the wave frequency.
By substituting the frequency \( f = 20.0 \text{ Hz} \), the angular frequency \( \omega \) is calculated as:\[ \omega = 2\pi \times 20.0 = 40\pi \text{ rad/s} \]This means the wave completes \( 40\pi \) radians every second. Angular frequency is essential in wave studies as it provides a more precise measure of wave oscillation for calculations involving complex wave interactions.
Wave Number
The wave number \( k \) defines the number of wave cycles in a unit distance and is essential for understanding wave propagation. The formula for determining the wave number is:
  • Formula: \( k = \frac{2\pi}{\lambda} \)
  • \( \lambda \) is the wavelength.
Given \( \lambda = 1.80 \text{ m} \), the wave number is:\[ k = \frac{2\pi}{1.80} = \frac{10\pi}{9} \text{ m}^{-1} \]This result indicates there are \( \frac{10\pi}{9} \) cycles per meter along the string. Understanding the wave number is important for analyzing spatial properties of waves, such as wave fronts and interference patterns.
Wave Function
The wave function describes how each point on a medium vibrates over time in a wave. For transverse waves like this one, the wave function is given by:
  • Formula: \( y(x, t) = A \sin(kx - \omega t + \phi) \)
  • \( A \) is the amplitude, \( k \) is the wave number, \( \omega \) is angular frequency, and \( \phi \) is phase constant.
In this context, the wave function becomes:\[ y(x, t) = 2.5 \sin\left(\frac{10\pi}{9}x - 40\pi t + \frac{\pi}{2}\right) \]This expression shows the displacement of any point at a location \( x \) at a time \( t \). It's a powerful tool to predict wave behavior across different regions of the string, crucial for various applications like generating and interpreting wave patterns.
Transverse Velocity
Transverse velocity describes how fast a single point on the string moves up and down as the wave travels along it. The maximum transverse velocity, \( v_{max} \), happens when the rate of change of wave function is greatest.
  • Formula: \( v_{max} = A \omega \)
  • \( A \) is amplitude and \( \omega \) is angular frequency.
Substituting the parameters \( A = 2.5 \text{ mm} = 0.0025 \text{ m} \) and \( \omega = 40\pi \text{ rad/s} \), we find:\[ v_{max} = 0.0025 \times 40\pi = 0.1\pi \text{ m/s} \]The transverse velocity is important in determining how kinetic energy is transferred through the medium. Understanding this helps in applications like acoustics and predicting how materials will respond to wave-induced vibrations.

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

You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^7\) m. You suspend a lead weight from the lower end of a light string that is 4.00 m long and has mass 0.0280 kg. You measure that it takes 0.0685 s for a transverse pulse to travel from the lower end to the upper end of the string. On the earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that you ignore its effect on the tension in the string. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

A musician tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \lambda/2\), (ii) \(x = \lambda /4\), and (iii) \(x = \lambda /8\), from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 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?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(Mg\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu\). You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M\); then you measure the node-to-node distance \(d\) for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

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