Harmonic Oscillator

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Problem:

Consider a harmonic oscillator of undamped frequency ω0 (= [itex]\sqrt{k/m}[/itex]) and damping constant β (=b/(2m), where b is the coefficient of the viscous resistance force).

a) Write the general solution for the motion of the position x(t) in terms of two arbitrary constants assuming an underdamped oscillator (ω0 > β).

b) Express the arbitrary constants in terms of the initial position of the oscillator x0 = x(t=0) and its initial speed v0 = [itex]\dot{x}[/itex](t=0).

c) Take the limit in the above solution where the oscillator becomes critically damped, β [itex]\rightarrow[/itex] ω0, assuming the initial conditions remain the same. (Be careful that β appears also in the constants of the solutions once they have been expressed in terms of initial conditions.)

d) Repeat steps (a) and (b) but now assuming an overdamped oscillator (β > ω0).

e) Take, again, the limit β [itex]\rightarrow[/itex] ω0 in the solutions found in (d). What do you conclude from the results (c) and (e)?


Solution:

a)
m[itex]\ddot{x}[/itex]=-b[itex]\dot{x}[/itex]-kx
m[itex]\ddot{x}[/itex]+b[itex]\dot{x}[/itex]+kx=0
[itex]\ddot{x}[/itex]+(b/m)[itex]\dot{x}[/itex]+(k/m)x=0
[itex]\ddot{x}[/itex]+2β[itex]\dot{x}[/itex]+(ω0)2=0
where β=b/(2m) and [itex]\sqrt{k/m}[/itex]=ω0
r=(-2β[itex]\pm[/itex](4β2-4ω02)1/2)/2
r=-β[itex]\pm[/itex](β202)1/2
r=-β[itex]\pm[/itex]iωD
where ωD2022
x(t)=e-βt(c1eDt+c2e-iωDt)
c1=Acos(δ)
c2=Asin(δ)
x(t)=Ae-βtcos(ωDt-δ) : General Solution

b)
[itex]\dot{x}[/itex](t)=-Ae-βtDsin(ωDt-δ)+βcos(ωDt-δ))
x0=Acos(δ)
[itex]\dot{x}[/itex]0=-A(βcos(δ)-ωDsin(δ))
x0β+[itex]\dot{x}[/itex]0=AωDsin(δ)
(x0β+[itex]\dot{x}[/itex]0)/x0Dtan(δ)
δ=arctan((x0β+[itex]\dot{x}[/itex]0)/(x0ωD))

I get stuck on part (c)
If I take the limit as β approaches ω0 then ωD goes to zero. And I get δ=∏/2. A turns out to be infinity. How can there be an infinite amplitude? Is this wrong?
 

Answers and Replies

  • #2
SteamKing
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Hint: look up resonance.
 
  • #3
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Correct me if I'm wrong but wouldn't resonance require an external driving force? This problem doesn't have an external driving force.
 
  • #4
vela
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Problem:

Consider a harmonic oscillator of undamped frequency ω0 (= [itex]\sqrt{k/m}[/itex]) and damping constant β (=b/(2m), where b is the coefficient of the viscous resistance force).

a) Write the general solution for the motion of the position x(t) in terms of two arbitrary constants assuming an underdamped oscillator (ω0 > β).

b) Express the arbitrary constants in terms of the initial position of the oscillator x0 = x(t=0) and its initial speed v0 = [itex]\dot{x}[/itex](t=0).

c) Take the limit in the above solution where the oscillator becomes critically damped, β [itex]\rightarrow[/itex] ω0, assuming the initial conditions remain the same. (Be careful that β appears also in the constants of the solutions once they have been expressed in terms of initial conditions.)

d) Repeat steps (a) and (b) but now assuming an overdamped oscillator (β > ω0).

e) Take, again, the limit β [itex]\rightarrow[/itex] ω0 in the solutions found in (d). What do you conclude from the results (c) and (e)?


Solution:

a)
m[itex]\ddot{x}[/itex]=-b[itex]\dot{x}[/itex]-kx
m[itex]\ddot{x}[/itex]+b[itex]\dot{x}[/itex]+kx=0
[itex]\ddot{x}[/itex]+(b/m)[itex]\dot{x}[/itex]+(k/m)x=0
[itex]\ddot{x}[/itex]+2β[itex]\dot{x}[/itex]+(ω0)2=0
where β=b/(2m) and [itex]\sqrt{k/m}[/itex]=ω0
r=(-2β[itex]\pm[/itex](4β2-4ω02)1/2)/2
r=-β[itex]\pm[/itex](β202)1/2
r=-β[itex]\pm[/itex]iωD
where ωD2022
x(t)=e-βt(c1eDt+c2e-iωDt)
c1=Acos(δ)
c2=Asin(δ)
x(t)=Ae-βtcos(ωDt-δ) : General Solution

b)
[itex]\dot{x}[/itex](t)=-Ae-βtDsin(ωDt-δ)+βcos(ωDt-δ))
x0=Acos(δ)
[itex]\dot{x}[/itex]0=-A(βcos(δ)-ωDsin(δ))
x0β+[itex]\dot{x}[/itex]0=AωDsin(δ)
(x0β+[itex]\dot{x}[/itex]0)/x0Dtan(δ)
δ=arctan((x0β+[itex]\dot{x}[/itex]0)/(x0ωD))

I get stuck on part (c)
If I take the limit as β approaches ω0 then ωD goes to zero. And I get δ=∏/2. A turns out to be infinity. How can there be an infinite amplitude? Is this wrong?
No, it's not wrong, but what's happening is a little obscure. Rewriting your solution slightly, you get
$$x(t) = e^{-\beta t}[(A\cos \delta)\cos \omega_D t + (A\sin\delta)\sin \omega_D t].$$ When ##\beta \to \omega_0##, ##A## indeed blows up and ##\delta \to \pi/2##. The product ##A\cos\delta##, however, remains finite. Similarly, ##\sin\omega_D t \to 0## because ##\omega_D \to 0##, and the product ##A \sin\omega_D t## remains finite.

You'll find it more straightforward to work with the solution in the form ##x(t) = e^{-\beta t}(A \cos \omega_D t + B \sin \omega_D t)##.
 
Last edited:

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