/*! 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 16 The equation of a wave travellin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 16

The equation of a wave travelling on a stretched string is: $$ y=4 \sin 2 \pi\left(\frac{t}{0.02}-\frac{x}{100}\right) $$ Here \(x\) and \(y\) are in \(\mathrm{cm}\) and \(t\) is in second. The speed of wave is: (a) \(50 \mathrm{~m} / \mathrm{s}\) (b) \(40 \mathrm{~m} / \mathrm{s}\) (c) \(50 \mathrm{~cm} / \mathrm{s}\) (d) \(40 \mathrm{~cm} / \mathrm{s}\)

Short Answer

Expert verified
The speed of the wave is 50 m/s (option a).

Step by step solution

01

Identify the wave equation parameters

Observe the given wave equation: \( y = 4 \sin 2 \pi \left( \frac{t}{0.02} - \frac{x}{100} \right) \). Notice that it is in the form of a sinusoidal wave function \( y = A \sin (2\pi f t - \frac{2\pi}{\lambda} x) \). Here, \( A \) is the amplitude, \( f \) is the frequency, and \( \lambda \) is the wavelength. By comparing terms, identify that \( \frac{1}{0.02} = 50 \) gives the frequency \( f \), and \( \frac{1}{100} \) gives the reciprocal of the wavelength \( \frac{2\pi}{\lambda} \).
02

Determine the frequency and wavelength

From the comparison with the standard wave equation, we have that the frequency \( f = \frac{1}{0.02} = 50 \) Hz, and using the expression \( \frac{2\pi}{\lambda} = \frac{2\pi}{100} \), we determine \( \lambda = 100 \) cm.
03

Calculate the wave speed

The speed \( v \) of a wave is given by the product of its frequency and wavelength: \( v = f \times \lambda \). Substitute the values: \( v = 50 \times 100 = 5000 \) cm/sec. Convert this speed into meters per second for clarity: \( v = 50 \) m/s.
04

Validate the answer

The wave speed of 5000 cm/sec is equivalent to 50 m/s. Therefore, the correct option that matches the computed speed is option (a) \( 50 \) m/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 91影视!

Key Concepts

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

Wave equation
In physics, a wave equation captures the properties of a wave through a mathematical expression. This expression typically relates displacement, position, frequency, and wavelength through a sinusoidal function. In the given exercise, the wave equation is represented as: \[ y = 4 \sin 2\pi \left( \frac{t}{0.02} - \frac{x}{100} \right) \] This equation has important components:
  • Amplitude \(A\): The value 4 in front of the sine function indicates the maximum displacement of the wave.
  • Frequency \(f\): Frequency can be extracted from the term \(\frac{t}{0.02}\).
  • Wave vector: The term \(\frac{x}{100}\) helps us determine the wavelength of the wave.
Such equations are fundamental in describing the behavior of waves moving through various mediums. They reveal how a wave propagates over time and distance.
Frequency of wave
The frequency of a wave denotes the number of complete cycles the wave completes per unit time. It is measured in hertz (Hz), and a higher frequency indicates that more waves pass a point within a given time frame. In the standard form of the wave equation, the frequency component is revealed as \(f = \frac{1}{T}\), where \(T\) is the period of the wave. From our wave equation, \[ y = 4 \sin 2\pi \left( \frac{t}{0.02} - \frac{x}{100} \right) \] The term \(\frac{t}{0.02}\) leads us to determine the frequency to be \(f = \frac{1}{0.02} = 50\) Hz. Understanding frequency is crucial, as it influences how we perceive sound and energy transfer through waves.
Wavelength
Wavelength, symbolized by \(\lambda\), is the distance over which the wave鈥檚 shape repeats. It is essential for understanding the spatial length of a wave cycle and is inversely related to the frequency.In the given wave equation, to find the wavelength, we look into the term associated with the \(x\) component, which is \(\frac{x}{100}\). We can equate this term with \(\frac{2\pi}{\lambda}\), which reveals \[ \lambda = 100 \text{ cm} \] This means the wave repeats every 100 cm along the string. The larger the wavelength, the further apart the waves are. In wave mechanics, understanding wavelength is vital for comprehending patterns of interference and wave propagation.
Stretched string wave
A stretched string wave is a common physics concept dealing with waves traveling along a string under tension. Such systems are typically used to model musical instruments, where vibrations travel in waves along the string.Characteristics of stretched string waves include:
  • They often appear as sinusoidal waves, showing regular oscillations.
  • The tension in the string affects the speed and characteristics of the wave.
  • The wave speed is calculated using the formula \(v = f \times \lambda\), where \(v\) depends on frequency \(f\) and wavelength \(\lambda\).
The exercise鈥檚 wave is traveling on such a stretched string, highlighting an essential principle: how physicists and engineers calculate wave speed to understand the dynamics of movement along strings, cables, and similar structures.

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

The wave travels along a string whose equation is \(y=\frac{p^{3}}{p^{2}+(p x-q t)^{2}}\) where \(p=2\) unit and \(q=0.5\) unit. The direction of propagation of wave is : (a) along \(+y\) -axis (b) along \(-x\) -axis (c) along \(+x\) -axis (d) none of these

If a string is stretched with a weight \(4 \mathrm{~kg}\) then the fundamental frequency is equal to \(256 \mathrm{~Hz}\). What weight is needed to produce its octave ? (a) \(4 \mathrm{~kg}\) wt (b) \(12 \mathrm{~kg}\) wt (c) \(16 \mathrm{~kg} \mathrm{wt}\) (d) \(24 \mathrm{~kg}\) wt

Equation of a plane wave is given by \(4 \sin \frac{\pi}{4}\left[2 t+\frac{x}{\overline{6}}\right]\) The phase difference at any given instant of two particles 16 \(\mathrm{cm}\) apart is: (a) \(60^{\circ}\) (b) \(90^{\circ}\) (c) \(30^{\circ}\) (d) \(120^{\circ}\)

A wave of angular frequency \(\omega\) propagates so that a certain phase of oscillation moves along \(x\) -axis, \(y\) -axis and \(z\) -axis with speeds \(c_{1}, c_{2}\) and \(c_{3}\) respectively. The propagation constant \(k\) is: (a) \(\frac{\omega}{\sqrt{c_{1}^{2}+c_{2}^{2}+c_{3}^{2}}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})\) (b) \(\frac{\omega}{c_{1}} \hat{i}+\frac{\omega}{c_{2}} \hat{\hat{j}}+\frac{\omega}{c_{3}} \hat{\mathbf{k}}\) (c) \((\omega \hat{\mathbf{1}}+\omega \hat{\mathrm{j}}+\omega \hat{\mathbf{k}}) \frac{1}{c}\) (d) none of the above

In a sonometer experiment the bridges are separated by a fixed distance. If \(T^{\prime}\) is tension in slightly elastic wire, which emits a tone of frequency ' \(n^{\prime}\), then frequency of the tone emitted by the wire when tension is increased to \(4 \%\), is : (a) \(n\) (b) \(2 n\) (c) slightly greater than \(n\) (d) slightly less than \(2 n\)
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Turning Points in Physics

Read Explanation

Scientific Method Physics

Read Explanation
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.