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) Plot a graph for $$ y(x, t)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s}) t-(314 \mathrm{rad} / \mathrm{cm}) x] $$ versus \(x\) at \(t=0\) and at \(t=\frac{1}{480} \mathrm{s}\). From the plots determine the amplitude, wavelength, and speed of the wave. (b) For the same function, plot a graph of \(y(x, t)\) versus \(t\) at \(x=0\) and find the period of the vibration. Show that \(\lambda=v T.\)

Short Answer

Expert verified
The amplitude of the wave is 4.0 cm, the wavelength is approximately 0.02 m, the speed of the wave is approximately 1.2 m/s, and the period of the wave is approximately 0.0166 s.

Step by step solution

01

Identify the given function and its properties

The given wave function is: $$ y(x, t)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s}) t-(314 \mathrm{rad} / \mathrm{cm}) x] $$ From this function, we can already identify the amplitude \(A\) as 4.0 cm, the angular frequency \(\omega\) as 378 rad/s, and the wave number \(k\) as 314 rad/cm.
02

Plot the graph of the function versus \(x\) at \(t=0\) and \(t=\frac{1}{480}\mathrm{s}\)

First, let's plot the wave function at \(t=0\): $$ y_t=0(x)=(4.0 \mathrm{cm}) \sin (-314 \mathrm{rad} / \mathrm{cm}\cdot x) $$ Also, plot the wave function at \(t=\frac{1}{480}\mathrm{s}\): $$ y_t=1/480(x)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s})\left(\frac{1}{480} \mathrm{s}\right)-(314 \mathrm{rad} / \mathrm{cm})x] $$ It is important to use appropriate software to generate these plots, then observe the graphs and deduce properties of the wave from the plot.
03

Calculate the amplitude, wavelength, and speed of the wave

We already know the amplitude \(A=4.0\mathrm{cm}\) from the wave function. To find the wavelength, we can use the formula: $$\lambda = \frac{2\pi}{k}$$ where \(k = 314\ \mathrm{rad}/\mathrm{cm}\), so, $$\lambda =\frac{2\pi}{314\ \mathrm{rad}/\mathrm{cm}} \approx 0.02\ \mathrm{m}$$ To calculate the speed of the wave, we can use the formula: $$v=\frac{\omega}{k}$$ where \(\omega=378\ \mathrm{rad/s}\), so $$v=\frac{378\ \mathrm{rad/s}}{314\ \mathrm{rad}/\mathrm{cm}} \approx 1.2\ \mathrm{m/s}$$
04

Plot the graph of the function versus \(t\) at \(x=0\)

Now let's plot the wave function at \(x=0\): $$ y_x=0(t)=(4.0 \mathrm{cm}) \sin (378 \mathrm{rad} / \mathrm{s} \cdot t) $$ Again, using appropriate software to generate the plot, observe the graph to find the period of the wave.
05

Calculate the period of the vibration and verify that \(\lambda=vT\)

To find the period of the vibration, we can use the formula: $$T = \frac{2\pi}{\omega}$$ where \(\omega = 378\ \mathrm{rad/s}\), so $$T= \frac{2\pi}{378\ \mathrm{rad/s}} \approx 0.0166\ \mathrm{s}$$ Now to show that \(\lambda=vT\), we have found the following values: $$\lambda \approx 0.02\ \mathrm{m}$$ $$v \approx 1.2\ \mathrm{m/s}$$ $$T \approx 0.0166\ \mathrm{s}$$ Multiplying \(v\) and \(T\), we get: $$vT = (1.2\ \mathrm{m/s})(0.0166\ \mathrm{s}) = 0.02\ \mathrm{m} = \lambda$$ This confirms that \(\lambda = vT\).

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!

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 longitudinal wave has a wavelength of \(10 \mathrm{cm}\) and an amplitude of \(5.0 \mathrm{cm}\) and travels in the \(y\) -direction. The wave speed in this medium is \(80 \mathrm{cm} / \mathrm{s},\) (a) Describe the motion of a particle in the medium as the wave travels through the medium. (b) How would your answer differ if the wave were transverse instead?
A transverse wave on a string is described by the equation $y(x, t)=(2.20 \mathrm{cm}) \sin [(130 \mathrm{rad} / \mathrm{s}) t+(15 \mathrm{rad} / \mathrm{m}) x].$ (a) What is the maximum transverse speed of a point on the string? (b) What is the maximum transverse acceleration of a point on the string? (c) How fast does the wave move along the string? (d) Why is your answer to (c) different from the answer to (a)?
When the tension in a cord is \(75 \mathrm{N}\), the wave speed is $140 \mathrm{m} / \mathrm{s} .$ What is the linear mass density of the cord?
While testing speakers for a concert. Tomás sets up two speakers to produce sound waves at the same frequency, which is between \(100 \mathrm{Hz}\) and $150 \mathrm{Hz}$. The two speakers vibrate in phase with one another. He notices that when he listens at certain locations, the sound is very soft (a minimum intensity compared to nearby points). One such point is \(25.8 \mathrm{m}\) from one speaker and \(37.1 \mathrm{m}\) from the other. What are the possible frequencies of the sound waves coming from the speakers? (The speed of sound in air is \(343 \mathrm{m} / \mathrm{s} .\) )
Tension is maintained in a string by attaching one end to a wall and by hanging a \(2.20-\mathrm{kg}\) object from the other end of the string after it passes over a pulley that is \(2.00 \mathrm{m}\) from the wall. The string has a mass per unit length of \(3.55 \mathrm{mg} / \mathrm{m} .\) What is the fundamental frequency of this string?
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