bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 18, Problem 32PQ

(a)

To determine

The first two harmonic frequencies of vibration for the wire.

(a)

Expert Solution
Check Mark

Answer to Problem 32PQ

The first two harmonic frequencies of vibration for the wire are 14.8 Hz and 29.7 Hz .

Explanation of Solution

Write the expression for the natural frequencies of vibration of a wire fixed at both ends.

  fn=n(v2L)       n=1,2,3,...                                                                                       (I)

Here, fn is the frequency of nth harmonic, L is the length of the wire and v is the speed of wave.

Write the expression for the speed of wave in terms of tension in string and mass per unit length.

  v=Tμ                                                                                                                  (II)

Here, T is the tension in string and μ is the mass per unit length.

Write the equation for μ .

  μ=mL

Here, m is mass of the string.

Put the above equation in equation (II).

  v=Tm/L                                                                                                             (III)

Put equation (III) in (I) to get final expression for fn .

  fn=n2LTm/L                                                                                                       (IV)

Conclusion:

It is given that length of the wire is 1.50 m, mass of the wire is 25.0 g and the tension in the wire is 33.0 N .

The first two harmonics are obtained when n is taken values 1 and 2 .

Substitute 1 for n, 1.50 m for L, 33.0 N for T and 25.0 g for m in equation (IV) to find the first harmonic.

  f1=12(1.50 m)33.0 N25.0 g1 kg1000 g/1.50 m=14.8 Hz

Substitute 2 for n, 1.50 m for L, 33.0 N for T and 25.0 g for m in equation (IV) to find the second harmonic.

  f2=22(1.50 m)33.0 N25.0 g1 kg1000 g/1.50 m=29.7 Hz

Therefore, the first two harmonic frequencies of vibration for the wire are 14.8 Hz and 29.7 Hz .

(b)

To determine

The possible harmonic modes of vibration of the wire.

(b)

Expert Solution
Check Mark

Answer to Problem 32PQ

The possible harmonic modes of vibration of the wire are n=5 harmonic at 74 Hz, n=10 at 148 Hz, n=15 at 222 Hz, etc. .

Explanation of Solution

It is given that the wire has a mode at 30.0 cm from one end. This could be any mode that has the node at 30.0 cm . Take D=30.0 cm to be the distance between the adjacent nodes.

Write the expression for D .

  D=λ2

Here, λ is the wavelength and D is the distance between the adjacent nodes.

Rewrite the above equation for λ .

  λ=2D                                                                                                                    (V)

Write the expression for the wavelength of harmonics.

  λ=2Ln

Rewrite the above equation for n .

  n=2Lλ                                                                                                                  (VI)

Put equation (V) in equation (VI).

  n=2L2D=LD

Substitute 1.50 m for L and 30.0 cm for D in the above equation to find the value of n .

  n=1.50 m30.0 cm1 m100 cm=1.50 m0.300 m=5

Thus the mode corresponds to the 5th harmonic.

Substitute 5 for n, 1.50 m for L, 33.0 N for T and 25.0 g for m in equation (IV) to find the frequency of the 5th harmonic.

  f5=52(1.50 m)33.0 N25.0 g1 kg1000 g/1.50 m=74 Hz

There could be also an entire wavelength fitting between the nodes at the end at 30.0 cm in which case D=λ . Any half number of wavelengths could fit between these two points and still have nodes at both.

Write the expression for D .

  D=Nλ2

Here, N is any integer number of half wavelengths between the two points on the string.

Rearrange the above equation for λ .

  λ=2DN

Put the above equation in equation (VI).

  n=2L2D/N=2NL2D=NLD

Substitute 1.50 m for L and 30.0 cm for D in the above equation to find the expression for n .

  n=N1.50 m30.0 cm1 m100 cm=N1.50 m0.300 m=5N

The expression for n implies that the harmonics that satisfy this condition are the 5th,10th,15th and so on.

Conclusion:

Substitute 10 for n, 1.50 m for L, 33.0 N for T and 25.0 g for m in equation (IV) to find the frequency of the 10th harmonic.

  f10=102(1.50 m)33.0 N25.0 g1 kg1000 g/1.50 m=148 Hz

Substitute 15 for n, 1.50 m for L, 33.0 N for T and 25.0 g for m in equation (IV) to find the frequency of the 15th harmonic.

  f15=152(1.50 m)33.0 N25.0 g1 kg1000 g/1.50 m=222 Hz

Therefore, the possible harmonic modes of vibration of the wire are n=5 harmonic at 74 Hz, n=10 at 148 Hz, n=15 at 222 Hz, etc. .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
No chatgpt pls will upvote
No chatgpt pls will upvote
No chatgpt pls will upvote

Chapter 18 Solutions

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term

Ch. 18 - The wave function for a pulse on a rope is given...Ch. 18 - Prob. 7PQCh. 18 - Prob. 8PQCh. 18 - Prob. 9PQCh. 18 - Prob. 10PQCh. 18 - Prob. 11PQCh. 18 - Two speakers, facing each other and separated by a...Ch. 18 - Prob. 13PQCh. 18 - Prob. 14PQCh. 18 - Prob. 15PQCh. 18 - As in Figure P18.16, a simple harmonic oscillator...Ch. 18 - A standing wave on a string is described by the...Ch. 18 - The resultant wave from the interference of two...Ch. 18 - A standing transverse wave on a string of length...Ch. 18 - Prob. 20PQCh. 18 - Prob. 21PQCh. 18 - Prob. 22PQCh. 18 - Prob. 23PQCh. 18 - A violin string vibrates at 294 Hz when its full...Ch. 18 - Two successive harmonics on a string fixed at both...Ch. 18 - Prob. 26PQCh. 18 - When a string fixed at both ends resonates in its...Ch. 18 - Prob. 28PQCh. 18 - Prob. 29PQCh. 18 - A string fixed at both ends resonates in its...Ch. 18 - Prob. 31PQCh. 18 - Prob. 32PQCh. 18 - Prob. 33PQCh. 18 - If you touch the string in Problem 33 at an...Ch. 18 - A 0.530-g nylon guitar string 58.5 cm in length...Ch. 18 - Prob. 36PQCh. 18 - Prob. 37PQCh. 18 - A barrel organ is shown in Figure P18.38. Such...Ch. 18 - Prob. 39PQCh. 18 - Prob. 40PQCh. 18 - The Channel Tunnel, or Chunnel, stretches 37.9 km...Ch. 18 - Prob. 42PQCh. 18 - Prob. 43PQCh. 18 - Prob. 44PQCh. 18 - If the aluminum rod in Example 18.6 were free at...Ch. 18 - Prob. 46PQCh. 18 - Prob. 47PQCh. 18 - Prob. 48PQCh. 18 - Prob. 49PQCh. 18 - Prob. 50PQCh. 18 - Prob. 51PQCh. 18 - Prob. 52PQCh. 18 - Prob. 53PQCh. 18 - Dog whistles operate at frequencies above the...Ch. 18 - Prob. 55PQCh. 18 - Prob. 56PQCh. 18 - Prob. 57PQCh. 18 - Prob. 58PQCh. 18 - Prob. 59PQCh. 18 - Prob. 60PQCh. 18 - Prob. 61PQCh. 18 - Prob. 62PQCh. 18 - The functions y1=2(2x+5t)2+4andy2=2(2x5t3)2+4...Ch. 18 - Prob. 64PQCh. 18 - Prob. 65PQCh. 18 - Prob. 66PQCh. 18 - Prob. 67PQCh. 18 - Prob. 68PQCh. 18 - Two successive harmonic frequencies of vibration...Ch. 18 - Prob. 70PQCh. 18 - Prob. 71PQCh. 18 - Prob. 72PQCh. 18 - A pipe is observed to have a fundamental frequency...Ch. 18 - The wave function for a standing wave on a...Ch. 18 - Prob. 75PQCh. 18 - Prob. 76PQCh. 18 - Prob. 77PQCh. 18 - Prob. 78PQCh. 18 - Prob. 79PQCh. 18 - Prob. 80PQ
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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
SIMPLE HARMONIC MOTION (Physics Animation); Author: EarthPen;https://www.youtube.com/watch?v=XjkUcJkGd3Y;License: Standard YouTube License, CC-BY