<< Chapter < Page Chapter >> Page >

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} } {}} } {}

Questions & Answers

how do you get the 2/50
Abba Reply
number of sport play by 50 student construct discrete data
Aminu Reply
width of the frangebany leaves on how to write a introduction
Theresa Reply
Solve the mean of variance
Veronica Reply
Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... Step 2: Find each score's deviation from the mean. ... Step 3: Square each deviation from the mean. ... Step 4: Find the sum of squares. ... Step 5: Divide the sum of squares by n – 1 or N.
kenneth
what is error
Yakuba Reply
Is mistake done to something
Vutshila
Hy
anas
hy
What is the life teble
anas
hy
Jibrin
statistics is the analyzing of data
Tajudeen Reply
what is statics?
Zelalem Reply
how do you calculate mean
Gloria Reply
diveving the sum if all values
Shaynaynay
let A1,A2 and A3 events be independent,show that (A1)^c, (A2)^c and (A3)^c are independent?
Fisaye Reply
what is statistics
Akhisani Reply
data collected all over the world
Shaynaynay
construct a less than and more than table
Imad Reply
The sample of 16 students is taken. The average age in the sample was 22 years with astandard deviation of 6 years. Construct a 95% confidence interval for the age of the population.
Aschalew Reply
Bhartdarshan' is an internet-based travel agency wherein customer can see videos of the cities they plant to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400 a. what is the probability of getting more than 12,000 hits? b. what is the probability of getting fewer than 9,000 hits?
Akshay Reply
Bhartdarshan'is an internet-based travel agency wherein customer can see videos of the cities they plan to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits
Akshay
1
Bright
Sorry i want to learn more about this question
Bright
Someone help
Bright
a= 0.20233 b=0.3384
Sufiyan
a
Shaynaynay
How do I interpret level of significance?
Mohd Reply
It depends on your business problem or in Machine Learning you could use ROC- AUC cruve to decide the threshold value
Shivam
how skewness and kurtosis are used in statistics
Owen Reply
yes what is it
Taneeya
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics for ap® courses' conversation and receive update notifications?

Ask