Download Harmonic oscillator: SHM, damped and forced and more Lecture notes Classical Mechanics in PDF only on Docsity!
Harmonic oscillator
Periodicity is abundant in nature – cycle of seasons, tides, climate phenomena like El Nino or monsoon, comet appearance, ECG / heart activity, crankshaft in car engine and, of course, pendulum. x
t
x
t
Simple harmonic motion : special type of periodic or oscillatory motion. Objects performing SHM are considered as simple harmonic oscillator. SHO is used in physics in wide range of ways : I (^) Analyzing complicated oscillatory behaviors in terms of SHO. I (^) Almost always the first step to check any different formulation of classical mechanics (Lagrangian, Hamiltonian, Hamilton-Jacobi). I (^) Harmonic oscillator in quantum mechanics gives first opportunity to understand the so-called zero point energy. I (^) Harmonic oscillator is used to model radiation classically or phenomena in solid state physics.
Simple pendulum
Let’s start with simple pendulum – it is far from simple.
r
θ
θ l
mg
T
P
Position, velocity and acceleration of mass m in Cartesian and spherical coordinates with P as origin,
~r = l (ˆi sin θ − ˆj cos θ) = l ˆr ~^ ˙r = l θ˙ (ˆi cos θ + ˆj sin θ) = l θ˙ θˆ ¨~r = l θ¨ (ˆi cos θ + ˆj sin θ) − l θ˙^2 (ˆi sin θ − ˆj cos θ) = l ¨θ θˆ − l θ˙^2 ˆr
Pull of gravity being the only external force,
ˆr : ml θ˙^2 = T − mg cos θ θˆ : ml θ¨ = −mg sin θ
Equation of motion for pendulum can be determine from the total energy as it remains invariant throughout the motion,
E =
m
l θ˙
dE dt ⇒ θ¨ +
g l sin θ = 0.
Motion described by equation of motion for small angle sin θ ≈ θ is called the Simple Harmonic Motion.
Kinetic and potential energies for small angle oscillation of pendulum,
KE =
m(l θ˙)^2 =
m x˙^2
PE = mgl(1 − cos θ) ≈ mgl
θ^2 2
mglθ^2
m
( (^) g l
(lθ)^2 =
mω^2 (lθ)^2 =
kx^2
Essentially SHM is the motion arising out of force law F = −k x, the
Hooke’s law – mass subjected to restoring force.
Simple Harmonic Oscillator equation and its solution:
m¨x = −kx ¨x + (k/m)x = 0 Let ω^2 = k/m and x(t) ∼ eαt^ be the solution. Then, ˙x = αx and x¨ = α^2 x. When substituted in the above SHO equation, gives
(α^2 + ω^2 ) x = 0 ⇒ α = ±iω ⇒ x(t) = C 1 eiωt^ + C 2 e−iωt However, the more familiar form of SHO solution is, x(t) = A cos ωt + B sin ωt. Equivalence of two solutions of SHO can be seen using Euler’s formula
e±iθ^ = cos θ ± i sin θ → cos θ = e
iθ (^) +e−iθ 2 sin θ = e
iθ (^) −e−iθ 2 i Thus the solution of SHO can be re-written as,
x(t) = A
eiωt^ + e−iωt 2
+ B
eiωt^ − e−iωt 2 i
A + iB 2 eiωt^ +
A − iB 2 e−iωt
= C 1 eiωt^ + C 2 e−iωt
Lissajous Figures
Harmonic oscillators of varying amplitude, frequency and phase are superposed to analyze complex periodic motions. When two simple harmonic motions in mutually perpendicular direction are superposed we get Lissajous figures.
x = a sin(ω 1 t + φ 1 ) and y = b sin(ω 2 t + φ 2 )
This gives rise to complex figures. Let’s simplify by assuming ω 1 = ω 2 = ω and φ 1 = 0, φ 2 = φ,
x = a sin ωt, y = b sin(ωt + φ) y = b sin ωt cos φ + b cos ωt sin φ y b
x a
cos φ +
( (^) x a
sin φ
y 2 b^2
x^2 a^2
cos^2 φ −
2 xy ab
cos φ = sin^2 φ −
x^2 a^2
sin^2 φ
y 2 b^2
x^2 a^2
2 xy ab
cos φ = sin^2 φ
Case I. φ = 0 :
( (^) y b −^
x a
= 0 ⇒ y = ba x motion in a straight line in xy -plane with positive slope b/a.
Case II. φ = π/2 : y^
2 b^2 +^
x^2 a^2 = 1 motion is an ellipse with semi major and minor axis a, b. If a = b, it is a circle. The ellipse / circle is rotating clockwise.
Case III. φ = π :
( (^) y b +^
x a
= 0 ⇒ y = − ba x motion in a straight line in xy -plane with negative slope −b/a.
Case IV. φ = 3π/2 : y^
2 b^2 +^
x^2 a^2 = 1 motion is an ellipse with semi major and minor axis a, b. If a = b, it is a circle. The ellipse / circle is rotating anti-clockwise.
Under damped motion : For b < ω, the roots α 1 , 2 are complex and conjugate to each other,
α 1 = −b + i
ω^2 − b^2 , α 2 = −b − i
ω^2 − b^2 x(t) = C 1 e−bt+it
√ ω^2 −b^2 + C 2 e−bt−it √ ω^2 −b^2
= e−bt^
[
A cos(t
ω^2 − b^2 ) + B sin(t
ω^2 − b^2
]
= Ce−bt^ cos(t
ω^2 − b^2 + φ)
We find the SHO is oscillating with a modified frequency and all along the amplitude gets reduced by the factor e−bt^ as time goes by.
Over damped motion : For b > ω, the roots α 1 , 2 are real and negative
α 1 = −b +
b^2 − ω^2 , α 2 = −b −
b^2 − ω^2 x(t) = C 1 e−bt+t
√b (^2) −ω 2
√b (^2) −ω 2
= e−bt^
[
C 1 e
√b (^2) −ω (^2) t
√b (^2) −ω (^2) t ]
There is no oscillation at all! The motion just dies out without even oscillating. Note at large t, the both term goes to zero.
How the first term goes to zero at large time?
Periodic forced oscillations
To sustain damped oscillatory motion, apply external time-dependent force F (t) = a cos ωt. The equation of motion is
x¨ + 2b x˙ + ω 02 x = f cos ωt
Here ω 0 is the natural frequency of the oscillator i.e. ω^20 = k/m. ω is the frequency of the driven force with amplitude f = a/m. The above is an inhomogeneous 2nd^ linear order differential equation (with constant coefficients) and its solution has two parts : (a) homogeneous solution i.e. when F (t) = 0. We already know this. (b) particular solution when F (t) 6 = 0. To arrive at solution, re-write f cos ωt → feiωt^. Since eventually the oscillator will oscillate with driven frequency ω, ansatz is x = Aeiωt^. The solution will be real part of x.
A =
f (ω^20 − ω^2 ) + 2ibω
⇒ x =
(ω 02 − ω^2 ) − 2 ibω (ω^20 − ω^2 )^2 + 4b^2 ω^2
feiωt
Extract the real part to determine the solution.
Since eiωt^ = cos ωt + i sin ωt, real part is (ω 02 − ω^2 ) cos ωt + 2bω sin ωt. Let (ω 02 − ω^2 ) = A cos φ and 2bω = A sin φ, then
A =
(ω 02 − ω^2 )^2 + 4b^2 ω^2 , tan φ = 2bω/(ω 02 − ω^2 )
Thus the full solution of forced oscillator is
x = Ce−bt^ cos(
ω^2 − b^2 t + α) + f √ (ω 02 − ω^2 )^2 + 4b^2 ω^2
cos(ωt − φ)
The homogeneous solution goes to zero as t → ∞ is transient in nature and the particular solution is steady-state nature.
