Damped harmonic motion problem

  • Thread starter meher4real
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  • #1
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Homework Statement:
Determine the deviation at a time when the amplitude has
dropped to 1/5 of the original value.
Relevant Equations:
A=A0 e^-bt
Hi !
Problem :
y = 5 e^-0.25t sin (0.5.t) (m, s). Determine the deviation at a time when the amplitude has
dropped to 1/5 of the original value.
I tried with A=A0 e^-bt=5 e^-0.25t
- Do i need to determine the time here or recreate the deviation equation when A decreased ? I don't understand the meaning of the question .
 

Answers and Replies

  • #2
Delta2
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Since I know french is your mother language, I think you translated wrong the term deviation. I think the problem wants the moment in time where the amplitude has drop from 5 to 1.

what is the original value of the amplitude?
What is the value of amplitude after time t?
 
  • #3
Delta2
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Ah on second thought by deviation you might mean the value of y. Then you first have to determine the time ##t_0## that the amplitude drops from 5 to 1, and then find ##y(t_0)##.
 
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  • #4
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hehe i'm mixing french with english what gives a masterpiece.
Sorry for the late response, i'm tired cause of no sleep, i'll be back in 8hours from now to work on it.
 
  • #5
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Ah on second thought by deviation you might mean the value of y. Then you first have to determine the time ##t_0## that the amplitude drops from 5 to 1, and then find ##y(t_0)##.
Hi !
You mean that i need to determine the value of t0 from the equation
A=A0 e^-bt0
and then calculate y(t0) ?
 
  • #6
berkeman
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Hi !
You mean that i need to determine the value of t0 from the equation
A=A0 e^-bt0
and then calculate y(t0) ?
It would help a lot if you could start to learn how to post mathematical equations using LaTeX instead of using just the keyboard to try to write equations. There is a helpful "LaTeX Guide" link just below the Edit box that will get you started.

For example, what you wrote above would look like this (I think, assuming that I'm interpreting what you wrote correctly):
$$A(t) = A_0 e^{b_0 t}$$
and then you calculate ##y(t_0)##?

Also, I'm confused by the title of the thread referring to "damped harmonic motion", but I don't see any oscillating term in your equations. Can you clarify that? Thanks.
 
  • #7
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The problem is from damped harmonic motion course.
I'll learn how to do it, thank you
this is my results A(t)=A0e^{-0.25t} => 1=5e^{-0.25t} then t=6.43s
yf=5e^{-0.25t}sin(0.5\pi t)
yf=0.17ms-1
 
  • #8
Steve4Physics
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this is my results A(t)=A0e^{-0.25t} => 1=5e^{-0.25t} then t=6.43s
yf=5e^{-0.25t}sin(0.5\pi t)
yf=0.17ms-1
I think the English word you need is 'displacement' (not 'deviation').

For t, you have a rounding error. What is 6.43775 rounded to 3 significant figures?

I disagree with your answer (value and unit!) for displacement:
##y(6.43775) = 1 sin(0.5 \pi \times 6.43775) = [##not ##0.17## and not ##ms^{-1}##].

I think you have made two mistakes:
a) forgotten to switch your calculator from degree to radians
b) incorrectly rounded the incorrect answer! (0.17558 rounded to 2 significant figures is 0.18, not 0.17).
 
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  • #9
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Thank you for your detailed response.
so t=6.438s.
then y(6.438)=0.18 Rad
 
  • #10
Steve4Physics
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Thank you for your detailed response.
so t=6.438s.
then y(6.438)=0.18 Rad
No.

From Post #1, y is a displacement measured in metres. For example y could be the height of a mass bouncing up and down on a spring, with y=0 the equilibrium height. Positive values of y are above the equilibrium position; negative values of y are below the equilibrium position. So y is a value in metres, not radians.

Note that ##0.5 \times \pi \times 6.43775 = 10.112##. This is a value in radians.

When you work out sin(10.112) you must make sure your calculator is set to radians. You are not working out sin(10.112º).

The value of sine itself has no units. When you multiply the amplitude (1m) by sin(10.112), the answer must be in metres.
 
  • #12
haruspex
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The concept of amplitude in a decaying oscillation is somewhat theoretical. You can only observe the amplitude at the peaks and troughs of the sine function.
Note that they are not the same as the peaks and troughs in the displacement. If you plot the amplitude curve and displacement curve together the amplitude curve does not pass through the peaks of displacement. Rather it touches the displacement curve, tangentially, a little after each peak displacement.
 
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