Damped Oscillator: Motion, Turning Points, and Constants | Homework Solution

Now how would I solve for K?Yes I think so. 2*\pi/(\OmegaD)= 5 seconds. =\tauTherefore , \exp(-tau*K) = 3 Now how would I solve for K?You can take the log of both sides. However, I have a feeling you are not supposed to do it that way.You know that (2*pi)/\Omega_D = 5, and you know that the log of the ratio of successive maxima is 3. Hence, you can use the information to find K.[You don't actually need to know the value of omega_D in order to find K.]You can take the log of both sides. However
  • #1
Benzoate
422
0

Homework Statement


A damped oscillator satisfies the equation

x'' + 2Kx' + [tex]\Omega[/tex]^2 *(x)

where K and [tex]\Omega[/tex] are positive constants with K < [tex]\Omega[/tex] (underdamping).

i)At time t =0 the particle is released from rest at the point x=a . Show that the subsequent motion is given by

x=a*exp(KT)(cos([tex]\Omega[/tex]D*t) +K/([tex]\Omega[/tex]2)*sin([tex]\Omega[/tex]D*t)


where [tex]\Omega[/tex]D=([tex]\Omega[/tex]^2 - K^2)^1/2.

ii)Find all the turning points of the function x(t) and show that the rati of successive maximum values of x is e^(-2*[tex]\pi[/tex]*K/([tex]\Omega[/tex]D)

iii)a certain damped oscillator has mass 10 kg , period 5 seconds and successive maximum values of its displacement are in the ratio 3:1. Find the values of the spring and damping constants [tex]\alpha[/tex] and [tex]\beta[/tex].

Homework Equations





The Attempt at a Solution



I had no trouble with part i) so I will skipped directly to part ii and iii.

ii) Not sure how to calculate the turning points at x(t) and why taking the ratio of those turning points is significant.

iii) How would knowing that finding the successive maximum values of its displacement are in the ratio 3 :1 aid me in finding the values of the spring and damping constants?
 
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  • #2
(ii) HINT: Turning points are stationary points.

(iii) I'm not quite sure which constants you are referring to since there are no alphas or betas in x(t).
 
  • #3
Hootenanny said:
(ii) HINT: Turning points are stationary points.

(iii) I'm not quite sure which constants you are referring to since there are no alphas or betas in x(t).

ii) so does that mean x(t) and x'(t) equal zero. What are the turning points supposed to be?
The problems wants me to find the ratio of successive maximum values of x(t) . When they are talking about successive maximum values at x(t) , do they mean x(t) at certain times and when the velocity is zero?
iii) I'm not sure what they are talking about either. I think the spring constant is K and I think the omega D constant is the damping constant so I have no idea what alpha and beta are.
 
  • #4
Benzoate said:
ii) so does that mean x(t) and x'(t) equal zero. What are the turning points supposed to be?
As I said in my previous post, turning points are stationary points, which are maxima, minima and points of inflection. Stationary points are so-called because the function stops increasing/decreasing at this point, in other words the function's derivative vanishes. So you need to find the points were x'(t)=0. Note that it is not necessary that x(t) vanishes.
Benzoate said:
The problems wants me to find the ratio of successive maximum values of x(t) . When they are talking about successive maximum values at x(t) , do they mean x(t) at certain times and when the velocity is zero?
As I said above, x'(t) vanishes at maxima and minima. Hence, the points at which x'(t)=0 could either be maxima or minima, you should consider each point and decide which it is.
 
  • #5
Hootenanny said:
As I said in my previous post, turning points are stationary points, which are maxima, minima and points of inflection. Stationary points are so-called because the function stops increasing/decreasing at this point, in other words the function's derivative vanishes. So you need to find the points were x'(t)=0. Note that it is not necessary that x(t) vanishes.

As I said above, x'(t) vanishes at maxima and minima. Hence, the points at which x'(t)=0 could either be maxima or minima, you should consider each point and decide which it is.

For the part 3 of this problem , I know what alpha and Beta are:

alpha= m*([tex]\Omega[/tex]2)

beta= 2*m*K

tau=2*pi/([tex]\Omega[/tex]D)= 2*pi/([tex]\Omega[/tex]2-K2)1/2)

tau is the period which is 5 seconds
I know I need to know K and omega in order to find alpha and Beta; I am not sure how to calculate the spring constant or omega.
 
  • #6
Benzoate said:
For the part 3 of this problem , I know what alpha and Beta are:

alpha= m*([tex]\Omega[/tex]2)

beta= 2*m*K

tau=2*pi/([tex]\Omega[/tex]D)= 2*pi/([tex]\Omega[/tex]2-K2)1/2)

tau is the period which is 5 seconds
I know I need to know K and omega in order to find alpha and Beta; I am not sure how to calculate the spring constant or omega.
HINT: You know that each maxima of x(t) it only one third of the magnitude of the previous maxima.
 
  • #7
Hootenanny said:
HINT: You know that each maxima of x(t) it only one third of the magnitude of the previous maxima.

So would I write something like:

the 3:1 ratio of successive maximum displacement values implies:

3/1 = e^-2pi*3*K/[tex]\Omega[/tex]D/(e^-2pi*K/[tex]\Omega[/tex]D to calculate K? then I could calculate OMEGA since I would now know K and [tex]\Omega[/tex]D
 
  • #8
Benzoate said:
So would I write something like:

the 3:1 ratio of successive maximum displacement values implies:

3/1 = e^-2pi*3*K/[tex]\Omega[/tex]D/(e^-2pi*K/[tex]\Omega[/tex]D to calculate K? then I could calculate OMEGA since I would now know K and [tex]\Omega[/tex]D
You're on the right lines. However, notice that the quantity

[tex]\exp\left(-\frac{2\pi K}{\Omega_D}\right)[/tex]

Is the ratio of the consecutive maxima of x(t), this what you are asked to show in (ii). The question then tells you that this ratio is equal to 3:1, in other words,

[tex]\exp\left(-\frac{2\pi K}{\Omega_D}\right) = 3[/tex]

Do you follow?
 
  • #9
Hootenanny said:
You're on the right lines. However, notice that the quantity

[tex]\exp\left(-\frac{2\pi K}{\Omega_D}\right)[/tex]

Is the ratio of the consecutive maxima of x(t), this what you are asked to show in (ii). The question then tells you that this ratio is equal to 3:1, in other words,

[tex]\exp\left(-\frac{2\pi K}{\Omega_D}\right) = 3[/tex]

Do you follow?


Yes I think so. 2*[tex]\pi[/tex]/([tex]\Omega[/tex]D)= 5 seconds. =[tex]\tau[/tex]

Therefore , [tex]\exp(-tau*K) = 3[/tex]
 

FAQ: Damped Oscillator: Motion, Turning Points, and Constants | Homework Solution

What is a damped oscillator?

A damped oscillator is a physical system that exhibits oscillatory motion, such as a mass attached to a spring. However, unlike an ideal oscillator, a damped oscillator experiences a decrease in amplitude over time due to the presence of a damping force. This damping force can be caused by factors such as friction or air resistance.

What are turning points in a damped oscillator?

Turning points, also known as extrema, are the points in a damped oscillator's motion where the velocity changes direction. These points occur when the kinetic energy of the oscillator is equal to the potential energy.

How do you calculate the constants in a damped oscillator?

The constants in a damped oscillator, such as the damping coefficient and the natural frequency, can be calculated using the equation of motion for the oscillator. This equation takes into account the mass, spring constant, and damping force acting on the system.

How does damping affect the motion of a damped oscillator?

Damping has a significant effect on the motion of a damped oscillator. It causes the amplitude of the oscillations to decrease over time, eventually leading to the oscillator coming to rest. Damping also affects the frequency of the oscillations, causing them to decrease as well.

What are some real-life examples of damped oscillators?

Damped oscillators can be found in many real-life systems, such as a swinging pendulum, a car's suspension system, or a guitar string. In each of these examples, there is some form of damping force that causes the oscillations to decrease over time.

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