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  1. By equating the work done to the energy removed, solve for the distance d size 12{x} {} .
  2. The work done by the non-conservative forces equals the initial, stored elastic potential energy. Identify the correct equation to use:
    W nc = Δ KE + PE = PE el,f PE el,i = 1 2 k μ k mg k 2 X 2 . size 12{W rSub { size 8{"nc"} } =Δ left ("KE"+"PE" right )="PE" rSub { size 8{"el,f"} } - "PE" rSub { size 8{"el,i"} } = { {1} over {2} } k left ( left ( { {μ rSub { size 8{k} } ital "mg"} over {k} } right ) rSup { size 8{2} } - X rSup { size 8{2} } right )} {}
  3. Recall that W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "Fd"} {} .
  4. Enter the friction as f = μ k mg size 12{F=μ rSub { size 8{k} } ital "mg"} {} into W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "Fd"} {} , thus
    W nc = μ k mgd . size 12{W size 8{"nc"}=μ rSub { size 8{k} } ital "mgd"} {}
  5. Combine these two equations to find
    1 2 k μ k mg k 2 X 2 = μ k mgd . size 12{ { {1} over {2} } k left ( left ( { {μ rSub { size 8{k} } } over {k} } right ) rSup { size 8{2} } - X rSup { size 8{2} } right )= - μ rSub { size 8{k} } } {}
  6. Solve the equation for d size 12{x} {} :
    d = k 2 μ k mg ( X 2 ( μ k mg k ) 2 ) . size 12{d= { { { {1} over {2} } ital "kX" rSup { size 8{2} } } over {μ rSub { size 8{k} } } } } {}
  7. Enter the known values into the resulting equation:
    d = 50 . 0 N/m 2 0 . 0800 0 . 200 kg 9 . 80 m/s 2 0 . 100 m 2 0 . 0800 0 . 200 kg 9 . 80 m/s 2 50 . 0 N/m 2 . size 12{d= { {"50" "." 0" N/m"} over {2 left (0 "." "0800" right ) left (0 "." "200"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} } left ( left (0 "." "100"" m" right ) rSup { size 8{2} } - left ( { { left (0 "." "0800" right ) left (0 "." "200"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} over {"50" "." 0" N/m"} } right ) rSup { size 8{2} } right )} {}
  8. Calculate d size 12{x} {} and convert units:
    d = 1 . 59 m . size 12{d=1 "." "59"`m} {}

Discussion b

This is the total distance traveled back and forth across x = 0 size 12{x=0} {} , which is the undamped equilibrium position. The number of oscillations about the equilibrium position will be more than d / X = ( 1 . 59 m ) / ( 0 . 100 m ) = 15 . 9 size 12{d/X= \( 1 "." "59"`m \) / \( 0 "." "100"`m \) ="15" "." 9} {} because the amplitude of the oscillations is decreasing with time. At the end of the motion, this system will not return to x = 0 size 12{x=0} {} for this type of damping force, because static friction will exceed the restoring force. This system is underdamped. In contrast, an overdamped system with a simple constant damping force would not cross the equilibrium position x = 0 size 12{x=0} {} a single time. For example, if this system had a damping force 20 times greater, it would only move 0.0484 m toward the equilibrium position from its original 0.100-m position.

This worked example illustrates how to apply problem-solving strategies to situations that integrate the different concepts you have learned. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknowns using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life.

Why are completely undamped harmonic oscillators so rare?

Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator.

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Describe the difference between overdamping, underdamping, and critical damping.

An overdamped system moves slowly toward equilibrium. An underdamped system moves quickly to equilibrium, but will oscillate about the equilibrium point as it does so. A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium.

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Test prep for ap courses

The non-conservative damping force removes energy from a system in which form?

  1. Mechanical energy
  2. Electrical energy
  3. Thermal energy
  4. None of the above

(c)

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The time rate of change of mechanical energy for a damped oscillator is always:

  1. 0
  2. Negative
  3. Positive
  4. Undefined
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A 0.5-kg object is connected to a spring that undergoes oscillatory motion. There is friction between the object and the surface it is kept on given by coefficient of friction μ k = 0.06 . If the object is released 0.2 m from equilibrium, what is the distance that the object travels? Given that the force constant of the spring is 50 N m -1 and the frictional force between the objects is 0.294 N.

d = k 2 μ K m g ( X 2 μ K m g k ) 2 w h e r e k = 50 N m 1 μ k = 0.06 m = 0.5 kg d = 50 N m 1 2 × 0.06 × 9.8 m s 2 ( ( 0.2 ) 2 ( ( 0.06 × 0.5 kg × 9.8 m s 2 ) 2 ( 50 N m 1 ) 2 ) ) = 1.698 m

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Section summary

  • Damped harmonic oscillators have non-conservative forces that dissipate their energy.
  • Critical damping returns the system to equilibrium as fast as possible without overshooting.
  • An underdamped system will oscillate through the equilibrium position.
  • An overdamped system moves more slowly toward equilibrium than one that is critically damped.

Conceptual questions

Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.)

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How would a car bounce after a bump under each of these conditions?

  • overdamping
  • underdamping
  • critical damping
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Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics?

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Problems&Exercises

The amplitude of a lightly damped oscillator decreases by 3 . 0% size 12{3 "." 0%} {} during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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