Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 61

Consider a guitar string stretching \(80.0 \mathrm{~cm}\) between its anchored ends. The string is tuned to play middle \(\mathrm{C},\) with a frequency of \(256 \mathrm{~Hz}\), when oscillating in its fundamental mode, that is, with one antinode between the ends. If the string is displaced \(2.00 \mathrm{~mm}\) at its midpoint and released to produce this note, what are the wave speed, \(v\), and the maximum speed, \(V_{\text {max }}\), of the midpoint of the string?

Short Answer

Expert verified
Answer: The wave speed is approximately 409.6 m/s, and the maximum speed of the midpoint of the string is approximately 10.05 m/s.

Step by step solution

01

Find the wavelength of the fundamental mode

In the fundamental mode, the length of the string is equal to half the wavelength. So, to find the wavelength \(λ\), we can use the relationship: \(λ = 2L\) Where \(L\) is the length of the string. We are given the length of the string, \(L = 80.0 \mathrm{~cm}\), so we can calculate the wavelength: \(λ = 2 \times 80.0 \mathrm{~cm} = 160.0 \mathrm{~cm} = 1.6 \mathrm{~m}\)
02

Calculate the wave speed

Now that we have the wavelength, we can calculate the wave speed using the formula: \(v = fλ\) We are given the frequency \(f = 256 \mathrm{~Hz}\) and we have found the wavelength \(λ = 1.6 \mathrm{~m}\), so we can calculate the wave speed: \(v = 256 \mathrm{~Hz} \times 1.6 \mathrm{~m} = 409.6 \mathrm{~m/s}\)
03

Calculate the angular frequency

To find the maximum speed of the midpoint, we need the angular frequency \(ω\). We can find it using the formula: \(ω = 2πf\) Where \(f\) is the frequency. We are given \(f = 256 \mathrm{~Hz}\), so we can calculate the angular frequency: \(ω = 2π \times 256 \mathrm{~Hz} = 512π \mathrm{~rad/s}\)
04

Find the maximum speed of the midpoint

Now we can find the maximum speed of the midpoint using the formula: \(V_{max} = Aω\) Where \(A\) is the amplitude and \(ω\) is the angular frequency. We are given the displacement of the midpoint, \(A = 2.00 \mathrm{~mm} = 0.002 \mathrm{~m}\), and we have found the angular frequency, \(ω = 512π \mathrm{~rad/s}\). We can calculate the maximum speed: \(V_{max} = 0.002 \mathrm{~m} \times 512π \mathrm{~rad/s} \approx 3.200π \mathrm{~m/s} \approx 10.05 \mathrm{~m/s}\) To summarize the results: - The wave speed, \(v\), is approximately \(409.6 \mathrm{~m/s}\). - The maximum speed, \(V_{max}\), of the midpoint of the string is approximately \(10.05 \mathrm{~m/s}\).

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.

Start your free trial

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 \(3.00-\mathrm{m}\) -long string, fixed at both ends, has a mass of \(6.00 \mathrm{~g}\). If you want to set up a standing wave in this string having a frequency of \(300 . \mathrm{Hz}\) and three antinodes, what tension should you put the string under?

The middle-C key (key 52 ) on a piano corresponds to a fundamental frequency of about \(262 \mathrm{~Hz},\) and the sopranoC key (key 64) corresponds to a fundamental frequency of \(1046.5 \mathrm{~Hz}\). If the strings used for both keys are identical in density and length, determine the ratio of the tensions in the two strings.

Hiking in the mountains, you shout "hey," wait \(2.00 \mathrm{~s}\) and shout again. What is the distance between the sound waves you cause? If you hear the first echo after \(5.00 \mathrm{~s}\), what is the distance between you and the point where your voice hit a mountain?

Suppose that the tension is doubled for a string on which a standing wave is propagated. How will the velocity of the standing wave change? a) It will double. c) It will be multiplied by \(\sqrt{2}\). b) It will quadruple. d) It will be multiplied by \(\frac{1}{2}\).

A wave traveling on a string has the equation of motion \(y(x, t)=0.02 \sin (5.00 x-8.00 t)\) a) Calculate the wavelength and the frequency of the wave. b) Calculate its velocity. c) If the linear mass density of the string is \(\mu=0.10 \mathrm{~kg} / \mathrm{m}\), what is the tension on the string?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electromagnetism

Read Explanation

Energy Physics

Read Explanation

Nuclear Physics

Read Explanation

Fields in Physics

Read Explanation

Electric Charge Field and Potential

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.

Sign-up for free