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 1.50-m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 62.0 m/s. What are the wavelength and frequency of (a) the fundamental; (b) the second overtone; (c) the fourth harmonic?

Short Answer

Expert verified
The fundamental has \( \lambda = 3.00 \) m and \( f \approx 20.67 \) Hz; the second overtone has \( \lambda = 1.00 \) m and \( f = 62.0 \) Hz; the fourth harmonic has \( \lambda = 0.75 \) m and \( f \approx 82.67 \) Hz.

Step by step solution

01

Understanding Wave Properties on a String

For a string fixed at both ends, the wavelengths of the standing waves are given by \( \lambda_n = \frac{2L}{n} \), where \( L = 1.50 \) m is the length of the string, and \( n \) is the harmonic number. The speed \( v \) of waves on the string is 62.0 m/s.
02

Calculating Wavelength for the Fundamental Frequency

The fundamental frequency, also known as the first harmonic, corresponds to \( n = 1 \). Use the formula \( \lambda_1 = \frac{2 \times 1.50}{1} = 3.00 \text{ m} \).
03

Finding Fundamental Frequency from Wave Speed and Wavelength

The frequency \( f_n \) of a wave is given by \( f_n = \frac{v}{\lambda_n} \). For the fundamental frequency: \( f_1 = \frac{62.0}{3.00} \approx 20.67 \text{ Hz} \).
04

Calculating Wavelength for the Second Overtone

The second overtone is the third harmonic, which corresponds to \( n = 3 \). Use the formula \( \lambda_3 = \frac{2 \times 1.50}{3} = 1.00 \text{ m} \).
05

Finding Frequency for the Second Overtone

For the second overtone: \( f_3 = \frac{62.0}{1.00} = 62.0 \text{ Hz} \).
06

Calculating Wavelength for the Fourth Harmonic

The fourth harmonic corresponds to \( n = 4 \). Use the formula \( \lambda_4 = \frac{2 \times 1.50}{4} = 0.75 \text{ m} \).
07

Finding Frequency for the Fourth Harmonic

For the fourth harmonic: \( f_4 = \frac{62.0}{0.75} \approx 82.67 \text{ Hz} \).

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.

Fundamental Frequency
In the context of waves on a string, the fundamental frequency is the simplest form of vibration, where the string has the least amount of nodes and antinodes required to sustain a wave. This is also referred to as the first harmonic. The fundamental frequency is crucial because it sets the baseline for understanding other harmonics or overtones in a wave system. It is represented when the string completes exactly one half of a wavelength within the span of the string length.
To calculate the fundamental frequency (\(f_1\)), you use the equation \(f_1 = \frac{v}{\lambda_1}\). Here, \(v\) is the wave speed, and \(\lambda_1\) is the wavelength of the fundamental frequency. This calculation helps us understand how fast the wave oscillates at its simplest harmonic state.
Harmonics
Harmonics refer to the multiple frequencies at which a system, such as a string, naturally oscillates. Beyond the fundamental frequency, there exist other harmonics that make the string vibrate in more complex patterns. These are the 2nd, 3rd, etc., harmonics, also called overtones.
The general expression for the wavelength of harmonics in a string fixed at both ends is given by \(\lambda_n = \frac{2L}{n}\), where \(L\) is the length of the string, and \(n\) is the harmonic number. Harmonics are integral because they embody the different modes of vibration a string can possess. Each mode comes with its own unique wavelength and frequency. For example, the third harmonic has three half-wavelength segments in the length of the string, leading to a different wavelength and frequency than the fundamental.
Wave Speed
Wave speed is an essential property that defines how quickly a wave travels through a medium. In this problem, we consider transverse waves on a string. The wave speed (\(v\)) depends on both the tension of the string and its linear density, although these specifics are not covered in our direct calculation.
To find the wave speed, the equation \(v = f \times \lambda\) is utilized, where \(f\) is the frequency and \(\lambda\) is the wavelength of the wave. This relation highlights the interdependence of wave properties and allows us to determine any one property if the others are known. In our problem, the speed is given by 62.0 m/s, helping us calculate frequencies and wavelengths of different harmonics.
Wave Properties
Wave properties encompass various aspects such as wavelength, frequency, amplitude, and speed. Each of these contributes to the behavior and characteristics of the wave.
  • Wavelength (\(\lambda\)): This is the distance over which the wave's shape repeats. Shorter wavelengths mean more waves are packed into a given space.
  • Frequency (\(f\)): This defines how many wave cycles pass a point in a given time period, usually a second. It is measured in hertz (Hz).
  • Amplitude: Though not directly covered in the problem, amplitude represents the wave's height and is related to the wave's energy.
  • Speed (\(v\)): This is the rate at which a wave propagates through a medium. Relating to all other properties through the formula \(v = f \times \lambda\), it underscores the interconnectedness of the elements.
Understanding these properties allows us to dissect and predict wave behavior in different mediums and situations.

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, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t) = (5.60 \, \mathrm{cm}) \mathrm{sin} [(0.0340 \mathrm{rad/cm})x] \mathrm{sin} [(150.0 \, \mathrm{rad/s})t]\), where the origin is at the left end of the string, the x-axis is along the string, and the \(y\)-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x,t)\) for this string if it were vibrating in its eighth harmonic?

A sinusoidal transverse wave travels on a string. The string has length 8.00 m and mass 6.00 g. The wave speed is 30.0 m/s, and the wavelength is 0.200 m. (a) If the wave is to have an average power of 50.0 W, what must be the amplitude of the wave? (b) For this same string, if the amplitude and wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?

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?

With what tension must a rope with length 2.50 m and mass 0.120 kg be stretched for transverse waves of frequency 40.0 Hz to have a wavelength of 0.750 m?

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 cm from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

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