Oscillations of Spring with Viscous Medium

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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
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on Phys.org
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?
 
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
 
I saw it ..they consider a for amplitude.However, I consider Xm
 
robax25 said:

The Attempt at a Solution



x=4.16cm[/B]
“The attempt at a solution” means your steps and thoughts. Please show effort.
 
Undamped SHM x(t)=Xmcos(ωt+∅)
 
for damped Oscillation x(t)=A e^-bt/2m cos(ωt+Φ). I did not notice it properly

Here A is amplitude
 
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.
 
how can I get the position at 3s?