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

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Homework Help Overview

The discussion revolves around a damped oscillator described by a second-order differential equation. The problem involves analyzing the motion of the oscillator, identifying turning points, and determining the spring and damping constants based on given conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the nature of turning points and their significance in the context of the motion described by x(t). Questions arise about how to identify these points and their relationship to the maxima of the function. There is also discussion about the constants involved in the equations and how they relate to the physical parameters of the system.

Discussion Status

Participants are actively engaging with the problem, raising questions about the definitions and implications of turning points. Some guidance has been provided regarding the relationship between the turning points and the derivatives of the function. There is ongoing exploration of how the ratio of successive maximum values relates to the constants in the equations.

Contextual Notes

There is a mention of specific values such as mass and period, as well as ratios of maximum displacements that are relevant to the calculations. Participants are also navigating the definitions of the constants involved, which are not explicitly defined in the problem statement.

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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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(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).
 
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.
 
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.
 
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.
 
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.
 
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
 
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?
 
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]
 

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