Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 16, Problem 84P

(a)

To determine

To Find:The maximum kinetic energy of the wire.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Length of the wire, l=2.00m

Tension in the wire, T=40.0N

Mass of the wire, m=0.100kg

At the midpoint, amplitude is A=2.00cm = 0.02 m

Formula Used:

Maximum kinetic energy of the wire can be obtained by:

  K.Emax=14mω2A2

Here, m is the mass, ω is the angular frequency and A is the amplitude of the wave.

  ω=2πf

Here, f is the frequency which can be obtained by:

  f=12lTμ

  μ=ml

Calculations:

Find the mass per unit length:

  μ=ml=0.100kg2.00m=0.05kg/m

Now calculate the frequency of the vibrating wire in fundamental mode:

  f=12lTμ=12×2.0040.00.1/2.00=7Hz

The angular frequency is:

  ω=2πf=2π(7.00)44rad/s

Now substitute all the known values to find the maximum kinetic energy of the wire:

  K.Emax=14mω2A2=14(0.100)(44rad/s)2(0.02m)2=0.0194J=19.4mJ

Conclusion:

Thus, the maximum kinetic energy of the wire is 19.4mJ .

(b)

To determine

To Find: The kinetic energy of the wire at the instant when transverse displacement is given by y=0.0200sin(π2x) .

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Length of the wire, l=2.00m

Tension in the wire, T=40.0N

Mass of the wire, m=0.100kg

At the midpoint, amplitude is A=2.00cm = 0.02 m

Displacement, y=0.0200sin(π2x)

  0.00mx2.00m

Formula Used:

Wave equation of standing wave in fundamental mode:

  y=Asin(kx)cos(ωt)

Calculations:

Compare the given displacement and the wave equation:

  y=0.0200sin(π2x) and y=Asin(kx)cos(ωt)

  cos(ωt)=1ω=0K.E=0

Conclusion:

Thus, the kinetic energy at the given instant would be zero.

(c)

To determine

To Find: The value of x for which the average value of the kinetic energy per unit length is the greatest.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Length of the wire, l=2.00m

Tension in the wire, T=40.0N

Mass of the wire, m=0.100kg

At the midpoint, amplitude is A=2.00cm = 0.02 m

Displacement, y=0.0200sin(π2x)

  0.00mx2.00m

Formula Used:

Average value of kinetic energy per unit length:

  dKdx=12μ(yt)2

Here, μ is the mass per unit length.

Wave equation of standing wave in fundamental mode:

  y=Asin(kx)cos(ωt)

Calculations:

  yt=Asinkx(ωsinωt)

For maxima, equate the derivative with zero.

  sin(kx)=0kx=0x=0orkx=πx=π2π/λx=λ2=(2l)2x=2(2.00)2=2.00m

Conclusion:

Thus, the value of x for which the average value of the kinetic energy per unit length is the greatest is 0.0m&2.00m .

(d)

To determine

To Find: The value of x for which the elastic potential energy per unit length has the maximum value.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

Length of the wire, l=2.00m

Tension in the wire, T=40.0N

Mass of the wire, m=0.100kg

At the midpoint, amplitude is A=2.00cm = 0.02 m

Displacement, y=0.0200sin(π2x)

  0.00mx2.00m

Formula Used:

Average value of elastic potential energy per unit length:

  dUdx=12μ(yx)2

Here, μ is the mass per unit length.

Wave equation of standing wave in fundamental mode:

  y=Asin(kx)cos(ωt)

Calculations:

  yx=Akcoskxcosωt

For maxima, equate the derivative with zero.

  cos(kx)=0kx=π2x=π/22π/λx=λ4=2l4x=2(2.00)4=1.0m

Conclusion:

Thus, the value of x for which the average value of the elastic potential energy per unit length is the greatest is at 1.0m .

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Chapter 16 Solutions

Physics for Scientists and Engineers, Vol. 1

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