EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106906149
Author: Jewett
Publisher: Cengage Learning US
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Chapter 16, Problem 51AP

(a)

To determine

The power P(x) carried by this wave at a point.

(a)

Expert Solution
Check Mark

Answer to Problem 51AP

The power P(x) carried by this wave at a point is 12kμω3A02e2bx.

Explanation of Solution

The wave function for string is given as,

  y(x,t)=(A0ebx)sin(kxωt)

Formula to calculate the velocity of wave for small segment of string is,

  v=dydt

Here, dy is the distance travel by wave in small segment and dt is the small time interval.

Substitute (A0ebx)sin(kxωt) for y.

  v=d(A0ebx)sin(kxωt)dtv=ω(A0ebx)cos(kxωt)

Assume the mass presents in the small segment dx of string is,

  dm=μdx

Here, μ is the linear mass density and dx is the small length of the string.

Formula to calculate the kinetic energy for small segment is,

    dK=12dmv2        (1)

Here, dK is the kinetic energy for small segment and dm is the mass of small segment of string.

Substitute μdx for m and ω(A0ebx)cos(kxωt) for v in equation (1).

    dK=12μdx(ω(A0ebx)cos(kxωt))2=12μω2(A0ebx)2cos2(kxωt)dx        (2)

Integrate the equation (2) over all the string elements in the wavelength of the waves for total kinetic energy.

    dK=0λ12μω2(A0ebx)2cos2(kxωt)dxK=14μω2(A0ebx)2λ

Formula to calculate the potential energy for string in small segment is,

    dp=12dmωy2        (3)

Here, dp is the potential energy present in the small segment of string.

Substitute μdx for m and (A0ebx)sin(kxωt) for y in equation (3).

    dp=12μdxω((A0ebx)sin(kxωt))2=12μdxω(A0ebx)2sin2(kxωt)        (4)

Integrate the equation (4) over all the string elements in the wavelength of the waves for total potential energy.

    dp=0λ12μω(A0ebx)2sin2(kxωt)dxp=14μω2(A0ebx)2λ

Formula to calculate the power P(x) is,

    P(x)=K+p        (5)

Substitute 14μω2(A0ebx)2λ for p and 14μω2(A0ebx)2λ for K in equation (5).

    P(x)=14μω2(A0ebx)2λ+14μω2(A0ebx)2λ=12μω2(A0ebx)2λ

Formula to calculate the wave length is,

    λ=ωk

Here, k is the wave number and λ is the wave number.

Substitute ωk for λ in equation (5) to get P(x).

    P(x)=12μω2(A0ebx)2ωk=12kμω3A02e2bx

Conclusion:

Therefore, the power P(x) carried by this wave at a point is 12kμω3A02e2bx.

(b)

To determine

The power P(0) carried by this wave at the origin.

(b)

Expert Solution
Check Mark

Answer to Problem 51AP

The power P(0) carried by this wave at the origin is 12kμω3A02.

Explanation of Solution

The wave function for string is given as,

    y(x,t)=(A0ebx)sin(kxωt)

Formula to calculate the power for wave in the string is,

    P(x)=12kμω2(A0ebx)2λ        (6)

Substitute 0 for x in equation (6).

    P(0)=12kμω2(A0eb×0)2λP(0)=12kμω2A02λ        (7)

Formula to calculate the wave length is,

    λ=ωk

Substitute ωk for λ in equation (7) to get P(0).

    P(0)=12kμω2A02ωk=12kμω3A02

Conclusion:

Therefore, the power P(0) carried by this wave at the origin is 12kμω3A02.

(c)

To determine

The ratio of P(x)P(0).

(c)

Expert Solution
Check Mark

Answer to Problem 51AP

The ratio of P(x)P(0) is e2bx.

Explanation of Solution

The wave function for string is given as,

    y(x,t)=(A0ebx)sin(kxωt)

The ratio of power given as,

    P(x)P(0)

Substitute 12kμω2A02 for P(0) and 12kμω2(A0ebx)2 for P(x).

    P(x)P(0)=12kμω3(A0ebx)212kμω3A02=e2bx

Conclusion:

Therefore, the ratio of P(x)P(0) is e2bx.

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Chapter 16 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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