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 Δ P = −( B ∂ s /∂ 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 ) = s max cos ( kx − ωt ). Show that this function satisfies the wave equation, provided ω / k = v = B / ρ .

Question

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Answer 1

Textbook Question

Chapter 16, Problem 60CP

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 = −(B ∂s/∂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) = s_max cos (kx − ωt). Show that this function satisfies the wave equation, provided ω / k = v = B / ρ .

(a)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Physics for Scientists and Engineers, Chapter 16, Problem 60CP , 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, Chapter 16, Problem 60CP , 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

To determine

The expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer to Problem 60CP

The expression −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 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)A−P(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.

∂F∂x=[−ΔP(x+Δx)+ΔP(x)]A=[−∂ΔP∂xx−∂ΔP∂xΔx+∂ΔP∂xx]A=−∂ΔP∂xΔxA

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=∂2s∂t2

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

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

∂F∂x=Δma (2)

Substitute ∂2s∂t2 for a, ρΔV for Δm and −∂ΔP∂xΔxA for ∂F∂x in equation (2).

−∂ΔP∂xΔxA=ρΔV∂2s∂t2

Conclusion:

Therefore the expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

(c)

Expert Solution

To determine

The wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer to Problem 60CP

The following wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Explanation of Solution

The value of the ΔP is B∂s∂x.

From part (b), the given expression is,

−∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2

Substitute B∂s∂x for ΔP.

−∂∂x(B∂s∂x)AΔx=ρAΔx∂2s∂t2Bρ∂2s∂x2=∂2s∂t2

Thus, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Conclusion:

Therefore, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

(d)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Explanation of Solution

The given wave equation is,

s(x,t)=smaxcos(kx−ωt) (3)

Apply the trial solution in the above equation.

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

∂s∂x=−ksmaxsin(kx−ωt)∂2s∂x2=−k2smaxcos(kx−ωt)

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

∂s∂t=+ωsmaxsin(kx−ωt)∂2s∂t2=−ωsmaxcos(kx−ωt)

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

Bρ∂2s∂x2=∂2s∂t2 (4)

Substitute −ωsmaxcos(kx−ωt) for ∂2s∂t2 and −k2smaxcos(kx−ωt) for ∂2s∂x2 in equation (2).

Bρ(−k2smaxcos(kx−ωt))=−ω2smaxcos(kx−ωt)Bρk2=ω2ωk=Bρωk=Bρ

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

Conclusion:

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

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A simple series circuit consists of a 150 Ω resistor, a 27.0 V battery, a switch, and a 2.00 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t =0s . Part A Part complete Part B Part complete Part C Find the electric flux at t =0.50ns. Express your answer in volt-meters. View Available Hint(s)for Part C Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type nothing V⋅m Part D Find the displacement current at t =0.50ns.

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Answer 2

Textbook Question

Chapter 16, Problem 60CP

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 = −(B ∂s/∂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) = s_max cos (kx − ωt). Show that this function satisfies the wave equation, provided ω / k = v = B / ρ .

(a)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Physics for Scientists and Engineers, Chapter 16, Problem 60CP , 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, Chapter 16, Problem 60CP , 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

To determine

The expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer to Problem 60CP

The expression −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 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)A−P(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.

∂F∂x=[−ΔP(x+Δx)+ΔP(x)]A=[−∂ΔP∂xx−∂ΔP∂xΔx+∂ΔP∂xx]A=−∂ΔP∂xΔxA

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=∂2s∂t2

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

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

∂F∂x=Δma (2)

Substitute ∂2s∂t2 for a, ρΔV for Δm and −∂ΔP∂xΔxA for ∂F∂x in equation (2).

−∂ΔP∂xΔxA=ρΔV∂2s∂t2

Conclusion:

Therefore the expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

(c)

Expert Solution

To determine

The wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer to Problem 60CP

The following wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Explanation of Solution

The value of the ΔP is B∂s∂x.

From part (b), the given expression is,

−∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2

Substitute B∂s∂x for ΔP.

−∂∂x(B∂s∂x)AΔx=ρAΔx∂2s∂t2Bρ∂2s∂x2=∂2s∂t2

Thus, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Conclusion:

Therefore, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

(d)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Explanation of Solution

The given wave equation is,

s(x,t)=smaxcos(kx−ωt) (3)

Apply the trial solution in the above equation.

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

∂s∂x=−ksmaxsin(kx−ωt)∂2s∂x2=−k2smaxcos(kx−ωt)

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

∂s∂t=+ωsmaxsin(kx−ωt)∂2s∂t2=−ωsmaxcos(kx−ωt)

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

Bρ∂2s∂x2=∂2s∂t2 (4)

Substitute −ωsmaxcos(kx−ωt) for ∂2s∂t2 and −k2smaxcos(kx−ωt) for ∂2s∂x2 in equation (2).

Bρ(−k2smaxcos(kx−ωt))=−ω2smaxcos(kx−ωt)Bρk2=ω2ωk=Bρωk=Bρ

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

Conclusion:

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

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Answer 3

Textbook Question

Chapter 16, Problem 60CP

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 = −(B ∂s/∂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) = s_max cos (kx − ωt). Show that this function satisfies the wave equation, provided ω / k = v = B / ρ .

(a)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Physics for Scientists and Engineers, Chapter 16, Problem 60CP , 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, Chapter 16, Problem 60CP , 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

To determine

The expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer to Problem 60CP

The expression −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 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)A−P(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.

∂F∂x=[−ΔP(x+Δx)+ΔP(x)]A=[−∂ΔP∂xx−∂ΔP∂xΔx+∂ΔP∂xx]A=−∂ΔP∂xΔxA

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=∂2s∂t2

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

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

∂F∂x=Δma (2)

Substitute ∂2s∂t2 for a, ρΔV for Δm and −∂ΔP∂xΔxA for ∂F∂x in equation (2).

−∂ΔP∂xΔxA=ρΔV∂2s∂t2

Conclusion:

Therefore the expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

(c)

Expert Solution

To determine

The wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer to Problem 60CP

The following wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Explanation of Solution

The value of the ΔP is B∂s∂x.

From part (b), the given expression is,

−∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2

Substitute B∂s∂x for ΔP.

−∂∂x(B∂s∂x)AΔx=ρAΔx∂2s∂t2Bρ∂2s∂x2=∂2s∂t2

Thus, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Conclusion:

Therefore, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

(d)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Explanation of Solution

The given wave equation is,

s(x,t)=smaxcos(kx−ωt) (3)

Apply the trial solution in the above equation.

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

∂s∂x=−ksmaxsin(kx−ωt)∂2s∂x2=−k2smaxcos(kx−ωt)

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

∂s∂t=+ωsmaxsin(kx−ωt)∂2s∂t2=−ωsmaxcos(kx−ωt)

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

Bρ∂2s∂x2=∂2s∂t2 (4)

Substitute −ωsmaxcos(kx−ωt) for ∂2s∂t2 and −k2smaxcos(kx−ωt) for ∂2s∂x2 in equation (2).

Bρ(−k2smaxcos(kx−ωt))=−ω2smaxcos(kx−ωt)Bρk2=ω2ωk=Bρωk=Bρ

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

Conclusion:

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

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Answer 4

Textbook Question

Chapter 16, Problem 60CP

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 = −(B ∂s/∂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) = s_max cos (kx − ωt). Show that this function satisfies the wave equation, provided ω / k = v = B / ρ .

(a)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Physics for Scientists and Engineers, Chapter 16, Problem 60CP , 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, Chapter 16, Problem 60CP , 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

To determine

The expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer to Problem 60CP

The expression −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 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)A−P(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.

∂F∂x=[−ΔP(x+Δx)+ΔP(x)]A=[−∂ΔP∂xx−∂ΔP∂xΔx+∂ΔP∂xx]A=−∂ΔP∂xΔxA

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=∂2s∂t2

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

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

∂F∂x=Δma (2)

Substitute ∂2s∂t2 for a, ρΔV for Δm and −∂ΔP∂xΔxA for ∂F∂x in equation (2).

−∂ΔP∂xΔxA=ρΔV∂2s∂t2

Conclusion:

Therefore the expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

(c)

Expert Solution

To determine

The wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer to Problem 60CP

The following wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Explanation of Solution

The value of the ΔP is B∂s∂x.

From part (b), the given expression is,

−∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2

Substitute B∂s∂x for ΔP.

−∂∂x(B∂s∂x)AΔx=ρAΔx∂2s∂t2Bρ∂2s∂x2=∂2s∂t2

Thus, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Conclusion:

Therefore, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

(d)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Explanation of Solution

The given wave equation is,

s(x,t)=smaxcos(kx−ωt) (3)

Apply the trial solution in the above equation.

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

∂s∂x=−ksmaxsin(kx−ωt)∂2s∂x2=−k2smaxcos(kx−ωt)

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

∂s∂t=+ωsmaxsin(kx−ωt)∂2s∂t2=−ωsmaxcos(kx−ωt)

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

Bρ∂2s∂x2=∂2s∂t2 (4)

Substitute −ωsmaxcos(kx−ωt) for ∂2s∂t2 and −k2smaxcos(kx−ωt) for ∂2s∂x2 in equation (2).

Bρ(−k2smaxcos(kx−ωt))=−ω2smaxcos(kx−ωt)Bρk2=ω2ωk=Bρωk=Bρ

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

Conclusion:

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

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Physics for Scientists and Engineers, Chapter 16, Problem 60CP , 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, Chapter 16, Problem 60CP , 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

To determine

The expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer to Problem 60CP

The expression −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 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)A−P(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.

∂F∂x=[−ΔP(x+Δx)+ΔP(x)]A=[−∂ΔP∂xx−∂ΔP∂xΔx+∂ΔP∂xx]A=−∂ΔP∂xΔxA

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=∂2s∂t2

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

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

∂F∂x=Δma (2)

Substitute ∂2s∂t2 for a, ρΔV for Δm and −∂ΔP∂xΔxA for ∂F∂x in equation (2).

−∂ΔP∂xΔxA=ρΔV∂2s∂t2

Conclusion:

Therefore the expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

(c)

Expert Solution

To determine

The wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer to Problem 60CP

The following wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Explanation of Solution

The value of the ΔP is B∂s∂x.

From part (b), the given expression is,

−∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2

Substitute B∂s∂x for ΔP.

−∂∂x(B∂s∂x)AΔx=ρAΔx∂2s∂t2Bρ∂2s∂x2=∂2s∂t2

Thus, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Conclusion:

Therefore, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

(d)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Explanation of Solution

The given wave equation is,

s(x,t)=smaxcos(kx−ωt) (3)

Apply the trial solution in the above equation.

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

∂s∂x=−ksmaxsin(kx−ωt)∂2s∂x2=−k2smaxcos(kx−ωt)

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

∂s∂t=+ωsmaxsin(kx−ωt)∂2s∂t2=−ωsmaxcos(kx−ωt)

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

Bρ∂2s∂x2=∂2s∂t2 (4)

Substitute −ωsmaxcos(kx−ωt) for ∂2s∂t2 and −k2smaxcos(kx−ωt) for ∂2s∂x2 in equation (2).

Bρ(−k2smaxcos(kx−ωt))=−ω2smaxcos(kx−ωt)Bρk2=ω2ωk=Bρωk=Bρ

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

Conclusion:

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

Answer 8

(a)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Physics for Scientists and Engineers, Chapter 16, Problem 60CP , 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, Chapter 16, Problem 60CP , 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.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

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

Answer 13

Answer to Problem 60CP

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

Physics for Scientists and Engineers, Chapter 16, Problem 60CP , additional homework tip 1

Answer 14

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, Chapter 16, Problem 60CP , 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.

Answer 15

(b)

Expert Solution

To determine

The expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer to Problem 60CP

The expression −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 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)A−P(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.

∂F∂x=[−ΔP(x+Δx)+ΔP(x)]A=[−∂ΔP∂xx−∂ΔP∂xΔx+∂ΔP∂xx]A=−∂ΔP∂xΔxA

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=∂2s∂t2

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

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

∂F∂x=Δma (2)

Substitute ∂2s∂t2 for a, ρΔV for Δm and −∂ΔP∂xΔxA for ∂F∂x in equation (2).

−∂ΔP∂xΔxA=ρΔV∂2s∂t2

Conclusion:

Therefore the expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer 20

Answer to Problem 60CP

The expression −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 is proved by applying the Newton’s second law to the element.

Answer 21

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)A−P(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.

∂F∂x=[−ΔP(x+Δx)+ΔP(x)]A=[−∂ΔP∂xx−∂ΔP∂xΔx+∂ΔP∂xx]A=−∂ΔP∂xΔxA

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=∂2s∂t2

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

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

∂F∂x=Δma (2)

Substitute ∂2s∂t2 for a, ρΔV for Δm and −∂ΔP∂xΔxA for ∂F∂x in equation (2).

−∂ΔP∂xΔxA=ρΔV∂2s∂t2

Conclusion:

Therefore the expression, −∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2 by applying the Newton’s second law to the element.

Answer 22

(c)

Expert Solution

To determine

The wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer to Problem 60CP

The following wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Explanation of Solution

The value of the ΔP is B∂s∂x.

From part (b), the given expression is,

−∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2

Substitute B∂s∂x for ΔP.

−∂∂x(B∂s∂x)AΔx=ρAΔx∂2s∂t2Bρ∂2s∂x2=∂2s∂t2

Thus, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Conclusion:

Therefore, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer 27

Answer to Problem 60CP

The following wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer 28

Explanation of Solution

The value of the ΔP is B∂s∂x.

From part (b), the given expression is,

−∂(ΔP)∂xAΔx=ρAΔx∂2s∂t2

Substitute B∂s∂x for ΔP.

−∂∂x(B∂s∂x)AΔx=ρAΔx∂2s∂t2Bρ∂2s∂x2=∂2s∂t2

Thus, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Conclusion:

Therefore, the wave equation for sound is Bρ∂2s∂x2=∂2s∂t2.

Answer 29

(d)

Expert Solution

To determine

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

Answer to Problem 60CP

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

Explanation of Solution

The given wave equation is,

s(x,t)=smaxcos(kx−ωt) (3)

Apply the trial solution in the above equation.

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

∂s∂x=−ksmaxsin(kx−ωt)∂2s∂x2=−k2smaxcos(kx−ωt)

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

∂s∂t=+ωsmaxsin(kx−ωt)∂2s∂t2=−ωsmaxcos(kx−ωt)

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

Bρ∂2s∂x2=∂2s∂t2 (4)

Substitute −ωsmaxcos(kx−ωt) for ∂2s∂t2 and −k2smaxcos(kx−ωt) for ∂2s∂x2 in equation (2).

Bρ(−k2smaxcos(kx−ωt))=−ω2smaxcos(kx−ωt)Bρk2=ω2ωk=Bρωk=Bρ

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

Conclusion:

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

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

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

Answer 34

Answer to Problem 60CP

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

Answer 35

Explanation of Solution

The given wave equation is,

s(x,t)=smaxcos(kx−ωt) (3)

Apply the trial solution in the above equation.

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

∂s∂x=−ksmaxsin(kx−ωt)∂2s∂x2=−k2smaxcos(kx−ωt)

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

∂s∂t=+ωsmaxsin(kx−ωt)∂2s∂t2=−ωsmaxcos(kx−ωt)

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

Bρ∂2s∂x2=∂2s∂t2 (4)

Substitute −ωsmaxcos(kx−ωt) for ∂2s∂t2 and −k2smaxcos(kx−ωt) for ∂2s∂x2 in equation (2).

Bρ(−k2smaxcos(kx−ωt))=−ω2smaxcos(kx−ωt)Bρk2=ω2ωk=Bρωk=Bρ

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

Conclusion:

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

Answer 36

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