Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
bartleby

Concept explainers

Question
Book Icon
Chapter 12, Problem 75GP

(a)

To determine

To Find: The wave speed and tension in the string

(a)

Expert Solution
Check Mark

Answer to Problem 75GP

Explanation of Solution

Given:

Length of violin

  l= 32 cm  = 0.32 m

Fundamental frequency f1=440 Hz

Mass per unit length  μ= 6.1×104kg/m

Formula used:

  v=f×2l m/s

Calculation:

Fundamental frequency has 2 nodes at the end and one anti-node in the middle, vibrating length of the string is 2l, so, the speed becomes

  v=f×2l m/s

  v=440×2×0.32 m/s

  v=281.6 m/s

The wave speed is 281.6 m/s.

Now, find the tension in the string, T

  T=v2μ N

  T=281.62 m/s×6.1×104 kg/m N

  T=48.37 N

The tension in the string is 48.37 N

Conclusion:

The wave speed and the tension in the string is 281.6 m/s and 48.37 N respectively.

(b)

To determine

To Find: The length of the tube of a simple wind instrument close at one end

(b)

Expert Solution
Check Mark

Answer to Problem 75GP

  L=0.195 m

Explanation of Solution

Given:

Fundamental frequency f1=440 Hz

Mass per unit length  μ= 6.1×104kg/m

Speed of the sound v= 343 m/s

Formula used:

  L=v4l m

Calculation:

The fundamental frequency is 440Hz .

The tube is closed, so the frequency calculated as

  L=v4l m

  L=343 m/s4×440Hz m

  L=0.1948 mL=0.195 m

The length of the tube is 0.195 m

Conclusion:

The length of the tube is 0.195 m of a simple wind instrument close at one end whose fundamental is also 440Hz .

(c)

To determine

To Find: The frequency of first overtone of each instrument.

(c)

Expert Solution
Check Mark

Answer to Problem 75GP

For Violin, f2=880 Hz 

For string, f2=1320 Hz

Explanation of Solution

Given:

Fundamental frequency f1=440 Hz

Speed of the sound v= 343 m/s

Formula used:

  f2=v2L×2 Hz

  f2=3×f1 Hz

Calculation:

To determine the frequency of first overtone at each instrument, use the following equations.

It is to be known that,

For Violin,

  f2=v2L×2 Hz

  f2=2×440 Hz 

  f2=880 Hz 

For String,

  f2=3×f1 Hz

  f2=3×440 Hz

  f2=1320 Hz

Conclusion:

The frequency of overtone of a violin and the string are 880 Hz, 1320 Hz respectively.

Chapter 12 Solutions

Physics: Principles with Applications

Ch. 12 - Prob. 11QCh. 12 - Prob. 12QCh. 12 - Traditional methods of protecting the hearing of...Ch. 12 - 14- Consider the two waves shown in Fig....Ch. 12 - Is there a Doppler shift if the source and...Ch. 12 - Prob. 16QCh. 12 - Prob. 17QCh. 12 - Prob. 1PCh. 12 - Prob. 2PCh. 12 - (a) Calculate the wavelengths in air at 20°C for...Ch. 12 - Prob. 4PCh. 12 - What is the intensity of a sound at the pain level...Ch. 12 - Prob. 6PCh. 12 - Prob. 7PCh. 12 - What is the sound level of a sound whose intensity...Ch. 12 - At a rock concert, a dB meter registered 130 dB...Ch. 12 - Prob. 10PCh. 12 - Prob. 11PCh. 12 - Prob. 12PCh. 12 - Prob. 13PCh. 12 - Prob. 14PCh. 12 - Prob. 15PCh. 12 - Prob. 16PCh. 12 - Prob. 17PCh. 12 - Prob. 18PCh. 12 - Prob. 19PCh. 12 - Prob. 20PCh. 12 - Prob. 21PCh. 12 - Prob. 22PCh. 12 - Prob. 23PCh. 12 - Prob. 24PCh. 12 - Prob. 25PCh. 12 - Prob. 26PCh. 12 - Prob. 27PCh. 12 - Prob. 28PCh. 12 - Prob. 29PCh. 12 - Prob. 30PCh. 12 - Prob. 31PCh. 12 - Prob. 32PCh. 12 - Prob. 33PCh. 12 - Prob. 34PCh. 12 - Prob. 35PCh. 12 - Prob. 36PCh. 12 - Prob. 37PCh. 12 - Prob. 38PCh. 12 - Prob. 39PCh. 12 - Prob. 40PCh. 12 - Prob. 41PCh. 12 - Prob. 42PCh. 12 - Prob. 43PCh. 12 - Prob. 44PCh. 12 - Prob. 45PCh. 12 - Prob. 46PCh. 12 - Prob. 47PCh. 12 - Prob. 48PCh. 12 - Prob. 49PCh. 12 - Prob. 50PCh. 12 - Prob. 51PCh. 12 - Prob. 52PCh. 12 - Prob. 53PCh. 12 - Prob. 54PCh. 12 - Prob. 55PCh. 12 - Prob. 56PCh. 12 - Prob. 57PCh. 12 - Prob. 58PCh. 12 - Prob. 59PCh. 12 - Prob. 60PCh. 12 - Prob. 61PCh. 12 - Prob. 62PCh. 12 - Prob. 63PCh. 12 - Prob. 64PCh. 12 - Prob. 65GPCh. 12 - Prob. 66GPCh. 12 - Prob. 67GPCh. 12 - Prob. 68GPCh. 12 - Prob. 69GPCh. 12 - Prob. 70GPCh. 12 - Prob. 71GPCh. 12 - Prob. 72GPCh. 12 - Prob. 73GPCh. 12 - Prob. 74GPCh. 12 - Prob. 75GPCh. 12 - Prob. 76GPCh. 12 - Prob. 77GPCh. 12 - Prob. 78GPCh. 12 - Prob. 79GPCh. 12 - Prob. 80GPCh. 12 - Prob. 81GPCh. 12 - Prob. 82GPCh. 12 - Prob. 83GPCh. 12 - Prob. 84GPCh. 12 - Prob. 85GPCh. 12 - Prob. 86GPCh. 12 - Prob. 87GPCh. 12 - Prob. 88GPCh. 12 - Prob. 89GPCh. 12 - Prob. 90GPCh. 12 - Prob. 91GPCh. 12 - Prob. 92GPCh. 12 - Prob. 93GP
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON