Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
9th Edition
ISBN: 9781305116429
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 17, Problem 17.72CP

In Section 16.7, we derived the speed of sound in a gas using the impulse–momentum theorem applied to the cylinder of gas in Figure 16.20. Let us find the speed of sound in a gas using a different approach based on the element of gas in Figure 16.18. Proceed as follows. (a) Draw a force diagram for this element showing the forces exerted on the left and right surfaces due to the pressure of the gas on either side of the element. (b) By applying Newton’s second law to the element, show that

( Δ P ) x A Δ x = ρ A Δ x 2 s t 2

(c) By substituting ΔP = −(Bs/∂x) (Eq. 16.30), derive the following wave equation for sound:

B ρ 2 s x 2 = 2 s t 2

(d) To a mathematical physicist, this equation demonstrates the existence of sound waves and determines their speed. As a physics student, you must take another step or two. Substitute into the wave equation the trial solution s(x, t) = smax cos (kxωt). Show that this function satisfies the wave equation, provided ω / k = v = B / ρ .

(a)

Expert Solution
Check Mark
To determine

The force diagram for this element showing the force exerted on the left and the right surface.

Answer to Problem 17.72CP

The force diagram for this element showing the force exerted on the left and the right surface is

Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics), Chapter 17, Problem 17.72CP , additional homework tip  1

Explanation of Solution

Force diagram contains all the forces acting on the body. It contains the direction of the each force acting on the body represents at its top and bottom end or left and right sides.

The force diagram for this element showing the force exerted on the left and the right surface is shown below.

Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics), Chapter 17, Problem 17.72CP , additional homework tip  2

Figure (1)

The force diagram of the element of gas in Figure (1) indicates the force exerted on the right and left surfaces due the pressure of the gas on the either side of the gas.

(b)

Expert Solution
Check Mark
To determine

The expression, (ΔP)xAΔx=ρAΔx2st2 by applying the Newton’s second law to the element.

Answer to Problem 17.72CP

The expression (ΔP)xAΔx=ρAΔx2st2 is proved by applying the Newton’s second law to the element.

Explanation of Solution

Let P(x) represents absolute pressure as a function of x.

The net force to the right on the chunk of air in Figure (1) is,

    +P(x)AP(x+Δx)A

The force due to atmosphere is,

    F=ΔP(x+Δx)A+ΔP(x)A        (1)

Here, A is the area of the surface, ΔP(x) is the atmospheric pressure on the surface and Δx is the smallest distance.

Differentiate the equation (1) with respect to x.

    Fx=[ΔP(x+Δx)+ΔP(x)]A=[ΔPxxΔPxΔx+ΔPxx]A=ΔPxΔxA

Formula to calculate the mass of the air is,

    Δm=ρΔV

Here, ρ is the density of the air, Δm is the mass of the air and ΔV is the volume of the air.

Formula to calculate the acceleration is,

    a=2st2

Here, s is the distance and t is the time.

From Newton’s second law, formula to calculate the Force is,

    Fx=Δma        (2)

Substitute 2st2 for a, ρΔV for Δm and ΔPxΔxA for Fx in equation (2).

    ΔPxΔxA=ρΔV2st2

Conclusion:

Therefore the expression, (ΔP)xAΔx=ρAΔx2st2 by applying the Newton’s second law to the element.

(c)

Expert Solution
Check Mark
To determine

The wave equation for sound is Bρ2sx2=2st2.

Answer to Problem 17.72CP

The following wave equation for sound is Bρ2sx2=2st2.

Explanation of Solution

 The value of the ΔP is Bsx.

From part (b), the given expression is,

    (ΔP)xAΔx=ρAΔx2st2

Substitute Bsx for ΔP.

    x(Bsx)AΔx=ρAΔx2st2Bρ2sx2=2st2

Thus, the wave equation for sound is Bρ2sx2=2st2.

Conclusion:

Therefore, the wave equation for sound is Bρ2sx2=2st2.

(d)

Expert Solution
Check Mark
To determine

The function s(x,t)=smaxcos(kxωt) satisfies the wave equation ωk=v=Bρ.

Answer to Problem 17.72CP

The function s(x,t)=smaxcos(kxωt) satisfies the wave equation ωk=v=Bρ.

Explanation of Solution

The given wave equation is,

    s(x,t)=smaxcos(kxωt)        (3)

Apply the trial solution in the above equation.

Double differentiate the equation (1) with respect to x.

    sx=ksmaxsin(kxωt)2sx2=k2smaxcos(kxωt)

Double differentiate the equation (1) with respect to t.

    st=+ωsmaxsin(kxωt)2st2=ωsmaxcos(kxωt)

The wave equation for sound in part (c) is,

    Bρ2sx2=2st2        (4)

Substitute ωsmaxcos(kxωt) for 2st2 and k2smaxcos(kxωt) for 2sx2 in equation (2).

    Bρ(k2smaxcos(kxωt))=ω2smaxcos(kxωt)Bρk2=ω2ωk=Bρωk=Bρ

Thus, the function s(x,t)=smaxcos(kxωt) satisfies the wave equation ωk=v=Bρ.

Conclusion:

Therefore, the function s(x,t)=smaxcos(kxωt) satisfies the wave equation ωk=v=Bρ.

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Chapter 17 Solutions

Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)

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