Oscillations of Spring with Viscous Medium

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Homework Help Overview

The discussion revolves around the oscillations of a spring-mass system in a viscous medium, focusing on the mathematical modeling of the displacement over time. The original poster presents a specific scenario involving given parameters such as spring constant, amplitude, and time intervals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expression for displacement, questioning the absence of an oscillating component in the original equation. There is confusion regarding the correct form of the equation for damped oscillations and the role of amplitude.

Discussion Status

Several participants are actively engaging in clarifying the mathematical representation of damped oscillations. Guidance has been offered regarding the need for a trigonometric factor in the displacement equation, and there is an ongoing exploration of how to derive the necessary constants for the equation.

Contextual Notes

Participants are navigating through assumptions about the amplitude and the damping factor, with some expressing uncertainty about the correct approach to incorporate trigonometric functions into their calculations. There is also mention of external resources for further clarification.

robax25
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Homework Statement


A spring with K=12N/m and an attached bob oscillates in a viscous medium.Amplitude is 6cm from equilibrium position at 1.5 s and Next amplitude of 5.6 cm occurs at 2.5s. what is its displacement at 3s and 4.5s and t=0s

Homework Equations



x(t)=Xme^-bt/2m

The Attempt at a Solution



x=4.16cm
[/B]
 

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Where is the oscillating portion of your x(t) expression?
 
e=0.133 as the ration of decay.

x(t)=xm x .133 and However, I am confused to find Xm(amplitude at 3 s) . I considered Xm=5.2cm at 3s. But, it is a false assumption.
 
Last edited:
robax25 said:
e=0.133 as the ration of decay.

x(t)=xm x .133 and However, I am confused to find Xm(amplitude at 3 s) . I considered Xm=5.2cm at 3s. But, it is a false assumption.
That is not what scottdave asked. The x(t) expression should have a trigonometric factor to represent the oscillation.
 
I do not get it. How can I calculate trigonometric factor?
 
robax25 said:
I do not get it. How can I calculate trigonometric factor?
You know the period. What is the general formula for a damped oscillation?
 
haruspex said:
You know the period. What is the general formula for a damped oscillation?
x(t)=Xme^-bt/2m here Xm is amplitude, b is damping factor
 
  • #10
robax25 said:
x(t)=Xme^-bt/2m here Xm is amplitude, b is damping factor
So is Xm the amplitude of the cosine wave? Did you look at the hyperphysics link I sent?
 
  • #11
I saw it ..they consider a for amplitude.However, I consider Xm
 
  • #12
robax25 said:

The Attempt at a Solution



x=4.16cm[/B]
“The attempt at a solution” means your steps and thoughts. Please show effort.
 
  • #13
robax25 said:
x(t)=Xme^-bt/2m here Xm is amplitude, b is damping factor
But clearly that does not produce an oscillation - unless that Xm is also a function of time.
What is the formula for an undamped SHM?
 
  • #14
Undamped SHM x(t)=Xmcos(ωt+∅)
 
  • #15
for damped Oscillation x(t)=A e^-bt/2m cos(ωt+Φ). I did not notice it properly

Here A is amplitude
 
  • #16
robax25 said:
for damped Oscillation x(t)=A e^-bt/2m cos(ωt+Φ). I did not notice it properly

Here A is amplitude
Right.
Now it is a matter of plugging the known facts into that equation to determine A, b/m, ω and φ.
So you need four equations. Knowing the displacement at a given time gives you one, and knowing that this is a local extrememum gives you another. You have that for two different times, giving you four equations altogether.
 
  • #17
how can I get the position at 3s?
 
  • #18
robax25 said:
how can I get the position at 3s?
Once you have determined the four constants as I described in post #16, you plug t=3 into your equation in post #15.
 

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