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Making connections: damped oscillator

Consider a mass attached to a spring. This system oscillates when in air because air exerts almost no force on the spring. Now put this system in a liquid, say, water. You will see that the system hardly oscillates when in water. When the system is submerged in water an external force acts on the oscillator. This force is exerted by the liquid against the motion of the spring-mass oscillator and is responsible for the inhibition of oscillations. A force that inhibits oscillations is called a “damping force,” and the system that experiences it is called a “damped oscillator." A damped oscillator sees a change in its energy. With time the total energy of the oscillator, which would be its mechanical energy, decreases. Since energy has to be conserved, the energy gets converted into thermal energy, which is stored in the water and the spring.

Damping an oscillatory motion: friction on an object connected to a spring

Damping oscillatory motion is important in many systems, and the ability to control the damping is even more so. This is generally attained using non-conservative forces such as the friction between surfaces, and viscosity for objects moving through fluids. The following example considers friction. Suppose a 0.200-kg object is connected to a spring as shown in [link] , but there is simple friction between the object and the surface, and the coefficient of friction μ k size 12{μ rSub { size 8{k} } } {} is equal to 0.0800. (a) What is the frictional force between the surfaces? (b) What total distance does the object travel if it is released 0.100 m from equilibrium, starting at v = 0 size 12{v=0} {} ? The force constant of the spring is k = 50 . 0 N/m size 12{k="50" "." 0`"N/m"} {} .

 The given figure (a) shows a spring on a frictionless surface attached to a bar or wall from the left side and on the right side of the spring, there is an object attached with mass m. Its amplitude is given by X, and X is equal to zero at the equilibrium level. Force F is applied to it from the right side, represented by a red arrow pointing toward the left and velocity v is equal to zero. An arrow showing the direction of force is also given alongside this figure as well as with the other four figures. The energy of the object is half k x squared.           In the given figure (b), after force is applied, the object moves to the left, compressing the spring slightly. The displacement of the object from its initial position is indicated by dots. The force F, here is equal to zero and velocity v, is maximum in the negative direction or the left. The energy of the object in this case is half m times negative v-max whole squared.           In the given figure (c), the spring has been compressed the maximum limit, and the amplitude is minus X. Now the force is toward the right, indicated here with a red arrow pointing to the right and the velocity, v, is zero. The energy of the object now is half k times negative x whole squared.           In the given figure (d), the spring is shown released from its compressed position and the object has moved toward the right side to reach the equilibrium level. Here, F is equal to zero, and the velocity, v, is the maximum. The energy of the object becomes half k times v max squared.           In the given figure (e), the spring has been stretched loose to the maximum possible limit and the object has moved to the far right. Now the velocity v, here is equal to zero and the direction of force is toward the left. As shown here, F is equal to zero. The energy of the object in this case is half k times x squared.
The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.

Strategy

This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. To solve an integrated concept problem, you must first identify the physical principles involved. Part (a) is about the frictional force. This is a topic involving the application of Newton’s Laws. Part (b) requires an understanding of work and conservation of energy, as well as some understanding of horizontal oscillatory systems.

Now that we have identified the principles we must apply in order to solve the problems, we need to identify the knowns and unknowns for each part of the question, as well as the quantity that is constant in Part (a) and Part (b) of the question.

Solution a

  1. Choose the proper equation: Friction is f = μ k mg size 12{F=μ rSub { size 8{k} } ital "mg"} {} .
  2. Identify the known values.
  3. Enter the known values into the equation:
    f = (0.0800) (0 .200 kg) (9 .80 m / s 2 ) . size 12{f=0 "." "0800" times 0 "." "200"`"kg" times 9 "." 8`"ms" rSup { size 8{"-2"} } } {}
  4. Calculate and convert units: f = 0.157 N . size 12{F=μ rSub { size 8{k} } ital "mg"} {}

Discussion a

The force here is small because the system and the coefficients are small.

Solution b

Identify the known:

  • The system involves elastic potential energy as the spring compresses and expands, friction that is related to the work done, and the kinetic energy as the body speeds up and slows down.
  • Energy is not conserved as the mass oscillates because friction is a non-conservative force.
  • The motion is horizontal, so gravitational potential energy does not need to be considered.
  • Because the motion starts from rest, the energy in the system is initially PE el,i = ( 1 / 2 ) kX 2 size 12{ ital "PE" rSub { size 8{e1} } = \( 1/2 \) ital "kX" rSup { size 8{2} } } {} . This energy is removed by work done by friction W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "fd"} {} , where d size 12{x} {} is the total distance traveled and f = μ k mg size 12{f=μk ital "mg"} {} is the force of friction. When the system stops moving, the friction force will balance the force exerted by the spring, so PE e1,f = ( 1 / 2 ) kx 2 size 12{"PE" rSub { size 8{"e1,f"} } = \( 1/2 \) ital "kx" rSup { size 8{2} } } {} where x size 12{x} {} is the final position and is given by
    F el = f kx = μ k mg x = μ k mg k . alignl { stack { size 12{F rSub { size 8{"el"} } =f} {} #ital "kx"=μ rSub { size 8{k} } ital "mg" {} # x= { {μ rSub { size 8{k} } ital "mg"} over {k} } {}} } {}

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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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