PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 15, Problem 103P

(a)

To determine

To calculate:The wave speed.

(a)

Expert Solution
Check Mark

Answer to Problem 103P

The wave speed (v) is 10.0m/s .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

The wave speed can be calculated as:

  v=FTμ

Where,

  v= Velocity of the wave.

  FT= tension in the wave.

  μ= Mass per unit length of the rope.

Calculation:

Let, the Δm represents the mass of the segment of the length Δx=1.00mm .

From the given data, wave speed can be found out as:

  v=FTμ

Where,

  v= Velocity of the wave.

  FT= tension in the wave.

  μ= Mass per unit length of the rope.

Now, substituting the all values of the linear density and tension in the equation:

  v= F T μv= 10.0N 0.100kg/mv=10.0m/s

Conclusion:

Thus, the wave speed (v) is 10.0m/s .

(b)

To determine

To calculate: The wavelength.

(b)

Expert Solution
Check Mark

Answer to Problem 103P

The wavelength (λ) is 2.00m .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

The wave speed can be calculated as:

  v= f λ

Where,

Sound’s speed: v .

Frequency of wave: .

The wavelength: λ .

Calculation:

To evaluate the wavelength, expression used is:

  v= f λ

Where,

Sound’s speed: v .

Frequency of wave: .

The wavelength: λ .

After substituting the values of the

Now, substituting the all values of the linear density and tension in the equation:

  λ=vλ=10.0m/s5.00s -1λ=2.00m

Conclusion:

Thus, the wavelength (λ) is 2.00m .

(c)

To determine

To calculate:The maximum transverse linear momentum.

(c)

Expert Solution
Check Mark

Answer to Problem 103P

The maximum transverse linear momentum (pmax) is 1.26×10-4kg.m/s .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

To calculate the maximum transverse linear momentum the expression used is:

  pmax=Δmvmax

Calculation:

Initially, relate the max transverse linear momentum of the 1.00mm segment to the max transverse speed of the wave.

  pmax=Δmvmaxpmax=μΔxAωpmax=2πfμΔxA

Substitute the numerical values to evaluate pmax:

  pmax=2πfμΔxApmax=2π(5.00s -1)(0.100kg/m)×(1 .00×10 -3m)(0.0400m)pmax=1.257×10-4kg.m/spmax=1.26×10-4kg.m/s

Conclusion:

Thus, the maximum transverse linear momentum (pmax) is 1.26×10-4kg.m/s .

(d)

To determine

To calculate: The maximum net force.

(d)

Expert Solution
Check Mark

Answer to Problem 103P

The maximum net force (Fmax) is 3.95mN .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

To calculate the maximum net forcethe expression used is:

  Fmax=Δmamax

Calculation:

The max net force acting on the segment is the product of the mass of th segment and its max acceleration:

  Fmax=ΔmamaxFmax=μΔxAω2Fmax=ωpmaxFmax=2πfpmax

Substitute the numerical values to evaluate Fmax :

  Fmax=2πfpmaxFmax=2π(5.00s -1)(1 .257×10 -4kg.m/s)Fmax=3.95mN

Conclusion:

Thus, the maximum net force (Fmax) is 3.95mN .

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Chapter 15 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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