Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 19, Problem 79PQ

(a)

To determine

The mass per unit length of the guitar string.

(a)

Expert Solution
Check Mark

Answer to Problem 79PQ

The mass per unit length of the guitar string is 3.95×104kg/m_.

Explanation of Solution

Write the expression for volume of the wire.

V=LA (I)

Here, V is the volume of the wire, L is the length of the wire, and A is the cross sectional area of the wire.

The wire has a circular cross section.

Write the expression for cross sectional area of wire.

A=πr2 (II)

Here, r is the radius of the wire.

Use equation (II) in (I).

V=L(πr2) (III)

Write the expression for linear mass density of the wire.

μ=mL (IV)

Here, μ is the linear mass density, and m is the mass of the guitar string.

Rewrite equation (I) to find L.

L=VA

Use above equation in (IV).

μ=AmV (V)

Use equation (II) in (V).

μ=πr2mV (VI)

Write the expression for density of the wire.

ρ=mV (VII)

Here, ρ is the density of the wire.

Use equation (VII) in (VI).

μ=πr2ρ . (VIII)

Write the expression to find radius of the wire.

r=d2 (IX)

Here, d is the diameter of the wire.

Conclusion:

Substitute 0.0254cm for d in equation (IX) to find r.

  r=0.0254cm×1m100cm2=1.27×104m

Substitute 1.27×104m for r, and 7.80×103kg/m3 for ρ in equation (VIII) to find μ.

  μ=π(1.27×104m)2(7.80×103kg/m3)=3.95×104kg/m

Therefore, the mass per unit length of the guitar string is 3.95×104kg/m_.

(b)

To determine

The tension in the guitar string.

(b)

Expert Solution
Check Mark

Answer to Problem 79PQ

The tension in the guitar string is 26.3N_.

Explanation of Solution

Write the expression for the fundamental frequency of vibration of the guitar string.

f1=v2L (X)

Here, f1 is the fundamental frequency of vibration of the string, and v is the speed of the waves.

Write the expression to find the speed of the waves in terms of the tension and mass density of the wire.

v=FTμ

Here, FT is the force of tension in the wire.

Use above expression in equation (X).

f1=12LFTμ (XI)

Solve equation (XI) for FT.

FT=μ(2Lf1)2 (XII)

Conclusion:

Substitute 3.95×104kg/m for μ, 65.78cm for L, and 196Hz for f1 in equation (XII) to find FT.

  FT=(3.95×104kg/m)[2(65.78cm×1m100cm)196Hz]2=26.3N

Therefore, the tension in the guitar string is 26.3N_.

(c)

To determine

The tension in the guitar string if the temperature is increased to 45.0°C.

(c)

Expert Solution
Check Mark

Answer to Problem 79PQ

The tension in the guitar string if the temperature is increased to 45.0°C is 23.5N_.

Explanation of Solution

Write the expression to find the unstressed length of the guitar string at 20.0°C.

L=Lnatural(1+FTAY) (XIII)

Here, L is the scale length of the guitar, Lnatural is the unstressed length of the guitar string at 20.0°C, Y is the Young’s Modulus of the material of the guitar string.

Solve equation (XIII) to find Lnatural.

Lnatural=L1+FT/AY (XIV)

Write the expression to find the unstressed length at 45.0°C.

L45°C=Lnatural[1+α(TfFi)] (XV)

Here, L45°C is the unstressed length of wire at 45.0°C, α linear expansion coefficient, Tf is the final temperature, and Ti is the initial temperature.

Write the equation to find the scale length of the string.

L=L45°C[1+FTAY] (XVI)

Here, FT is the force of tension at 45.0°C.

Solve equation (XVI) for FT.

FT=AY[LL45°C1] (XVII)

Conclusion:

Substitute 1.27×104m for r in equation (II) to find A.

  A=π(1.27×104m)2=5.06×108m2

Substitute 26.3N for FT, 65.78cm for L, 5.06×108m2 for A, and 20.0×1010Pa for Y in equation (XIV) to find Lnatural.

  Lnatural=65.78cm×1m100cm1+26.3N(5.06×108m2)(20.0×1010Pa)=0.6561m

Substitute 0.6561m for Lnatural, 11.0×106°C1 for α, 45.0°C for Tf, and 20.0°C for Ti in equation (XV) to find L45°C.

  L45°C=(0.6561m)[1+(11.0×106°C1)(45.0°C20.0°C)]=0.6563m

Substitute 5.06×108m2 for A, 20.0×1010Pa for Y, 0.6578m for L, and 0.6563m for L45°C in equation (XVII) to find FT.

  FT=(5.06×108m2)(20.0×1010Pa)[0.6578m0.6563m1]=23.5N

Therefore, the tension in the guitar string if the temperature is increased to 45.0°C is 23.5N_.

(d)

To determine

The fundamental frequency of guitar string at 45.0°C.

(d)

Expert Solution
Check Mark

Answer to Problem 79PQ

The fundamental frequency of guitar string at 45.0°C is 185Hz_.

Explanation of Solution

Write the expression for the ratio of frequency at 45.0°C and 20.0°C.

f1f1=FTFT (XVIII)

Here, f1 is the frequency at 45.0°C, and FT is the tension at 45.0°C.

Solve equation (XVIII) for f1.

f=f1FTFT (XIX)

Conclusion:

Substitute 196Hz for f1, 23.5N for FT, and 26.3N for FT in equation (XIX) to find f1.

  f1=(196Hz)23.5N26.3N=185Hz

Therefore, the fundamental frequency of guitar string at 45.0°C is 185Hz_.

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
2.62 Collision. The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track (Fig. P2.62). The freight train is traveling at 15.0 m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of 0.100 m/s² in a direction opposite to the train's velocity, while the freight train continues with constant speed. Take x = 0 at the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place? (c) On a single graph, sketch the positions of the front of the pas- senger train and the back of the freight train.
Can I get help with how to calculate total displacement? The answer is 78.3x-4.8y
2.70 Egg Drop. You are on the Figure P2.70 roof of the physics building, 46.0 m above the ground (Fig. P2.70). Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your profes- sor's head, where should the profes- sor be when you release the egg? Assume that the egg is in free fall. 2.71 CALC The acceleration of a particle is given by ax(t) = -2.00 m/s² +(3.00 m/s³)t. (a) Find the initial velocity Vox such that v = 1.20 m/s 1.80 m 46.0 m

Chapter 19 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

Ch. 19 - Prob. 4PQCh. 19 - Prob. 5PQCh. 19 - Prob. 6PQCh. 19 - Prob. 7PQCh. 19 - Prob. 8PQCh. 19 - Object A is placed in thermal contact with a very...Ch. 19 - Prob. 10PQCh. 19 - Prob. 11PQCh. 19 - Prob. 12PQCh. 19 - Prob. 13PQCh. 19 - The tallest building in Chicago is the Willis...Ch. 19 - Prob. 15PQCh. 19 - Prob. 16PQCh. 19 - At 22.0C, the radius of a solid aluminum sphere is...Ch. 19 - Prob. 18PQCh. 19 - Prob. 19PQCh. 19 - Prob. 20PQCh. 19 - The distance between telephone poles is 30.50 m in...Ch. 19 - Prob. 22PQCh. 19 - Prob. 23PQCh. 19 - Prob. 24PQCh. 19 - Prob. 25PQCh. 19 - Prob. 26PQCh. 19 - Prob. 27PQCh. 19 - Prob. 28PQCh. 19 - Prob. 29PQCh. 19 - Prob. 30PQCh. 19 - Prob. 31PQCh. 19 - Prob. 32PQCh. 19 - Prob. 33PQCh. 19 - Prob. 34PQCh. 19 - Prob. 35PQCh. 19 - Prob. 36PQCh. 19 - Prob. 37PQCh. 19 - Prob. 38PQCh. 19 - Prob. 39PQCh. 19 - On a hot summer day, the density of air at...Ch. 19 - Prob. 41PQCh. 19 - Prob. 42PQCh. 19 - Prob. 43PQCh. 19 - Prob. 44PQCh. 19 - Prob. 45PQCh. 19 - Prob. 46PQCh. 19 - Prob. 47PQCh. 19 - A triple-point cell such as the one shown in...Ch. 19 - An ideal gas is trapped inside a tube of uniform...Ch. 19 - Prob. 50PQCh. 19 - Prob. 51PQCh. 19 - Case Study When a constant-volume thermometer is...Ch. 19 - An air bubble starts rising from the bottom of a...Ch. 19 - Prob. 54PQCh. 19 - Prob. 55PQCh. 19 - Prob. 56PQCh. 19 - Prob. 57PQCh. 19 - Prob. 58PQCh. 19 - Prob. 59PQCh. 19 - Prob. 60PQCh. 19 - Prob. 61PQCh. 19 - Prob. 62PQCh. 19 - Prob. 63PQCh. 19 - Prob. 64PQCh. 19 - Prob. 65PQCh. 19 - Prob. 66PQCh. 19 - Prob. 67PQCh. 19 - Prob. 68PQCh. 19 - Prob. 69PQCh. 19 - Prob. 70PQCh. 19 - Prob. 71PQCh. 19 - A steel plate has a circular hole drilled in its...Ch. 19 - Prob. 73PQCh. 19 - A gas is in a container of volume V0 at pressure...Ch. 19 - Prob. 75PQCh. 19 - Prob. 76PQCh. 19 - Prob. 77PQCh. 19 - Prob. 78PQCh. 19 - Prob. 79PQCh. 19 - Prob. 80PQCh. 19 - Two glass bulbs of volumes 500 cm3 and 200 cm3 are...
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
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
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
Kinetic Molecular Theory and its Postulates; Author: Professor Dave Explains;https://www.youtube.com/watch?v=o3f_VJ87Df0;License: Standard YouTube License, CC-BY