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Harmonic Oscillator Problem: Consideration & Solutions
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[QUOTE="mattmatt, post: 4522934, member: 488222"] [U]Problem:[/U] Consider a harmonic oscillator of undamped frequency ω[SUB]0[/SUB] (= [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 (ω[SUB]0[/SUB] > β). b) Express the arbitrary constants in terms of the initial position of the oscillator x[SUB]0[/SUB] = x(t=0) and its initial speed v[SUB]0[/SUB] = [itex]\dot{x}[/itex](t=0). c) Take the limit in the above solution where the oscillator becomes critically damped, β [itex]\rightarrow[/itex] ω[SUB]0[/SUB], 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 > ω[SUB]0[/SUB]). e) Take, again, the limit β [itex]\rightarrow[/itex] ω[SUB]0[/SUB] in the solutions found in (d). What do you conclude from the results (c) and (e)?[U]Solution:[/U] 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]+(ω[SUB]0[/SUB])[SUP]2[/SUP]=0 where β=b/(2m) and [itex]\sqrt{k/m}[/itex]=ω[SUB]0[/SUB] r=(-2β[itex]\pm[/itex](4β[SUP]2[/SUP]-4ω[SUB]0[/SUB][SUP]2[/SUP])[SUP]1/2[/SUP])/2 r=-β[itex]\pm[/itex](β[SUP]2[/SUP]-ω[SUB]0[/SUB][SUP]2[/SUP])[SUP]1/2[/SUP] r=-β[itex]\pm[/itex]iω[SUB]D[/SUB] where ω[SUB]D[/SUB][SUP]2[/SUP]=ω[SUB]0[/SUB][SUP]2[/SUP]-β[SUP]2[/SUP] x(t)=e[SUP]-βt[/SUP](c[SUB]1[/SUB]e[SUP]iω[SUB]D[/SUB]t[/SUP]+c[SUB]2[/SUB]e[SUP]-iω[SUB]D[/SUB]t[/SUP]) c[SUB]1[/SUB]=Acos(δ) c[SUB]2[/SUB]=Asin(δ) x(t)=Ae[SUP]-βt[/SUP]cos(ω[SUB]D[/SUB]t-δ) : General Solution b) [itex]\dot{x}[/itex](t)=-Ae[SUP]-βt[/SUP](ω[SUB]D[/SUB]sin(ω[SUB]D[/SUB]t-δ)+βcos(ω[SUB]D[/SUB]t-δ)) x[SUB]0[/SUB]=Acos(δ) [itex]\dot{x}[/itex][SUB]0[/SUB]=-A(βcos(δ)-ω[SUB]D[/SUB]sin(δ)) x[SUB]0[/SUB]β+[itex]\dot{x}[/itex][SUB]0[/SUB]=Aω[SUB]D[/SUB]sin(δ) (x[SUB]0[/SUB]β+[itex]\dot{x}[/itex][SUB]0[/SUB])/x[SUB]0[/SUB]=ω[SUB]D[/SUB]tan(δ) δ=arctan((x[SUB]0[/SUB]β+[itex]\dot{x}[/itex][SUB]0[/SUB])/(x[SUB]0[/SUB]ω[SUB]D[/SUB])) I get stuck on part (c) If I take the limit as β approaches ω[SUB]0[/SUB] then ω[SUB]D[/SUB] goes to zero. And I get δ=∏/2. A turns out to be infinity. How can there be an infinite amplitude? Is this wrong? [/QUOTE]
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Harmonic Oscillator Problem: Consideration & Solutions
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