PHYS1110H, 2011 Fall.
Shijie Zhong
Harmonic Oscillator
1-D Harmonic oscillation
€
€
€
€
For the spring-block system, when the block only has the spring force acting
on it (i.e., the external force is removed, and no friction), the Newton’s
second law for the block is
d2x
.
(1)
−kx = ma = m
2
dt
2
d x k
(2)
+ x = 0.
2
m
dt
Equation 2 is a second order ordinary differential equation (ODE). If we
define frequency
k
ω=
,
(3)
m
and without getting to the details of solution procedures, we know that the
following solution for x satisfies equation 2.
(4)
x(t) = h sin(ωt + φ0 ) ,
where h is the maximum amplitude of the block’s displacement and φ0, both
of which can be determined by initial conditions of the block. For example,
in example 4, at t=0, the displacement x=-h, φ0 must be equal to –π/2. The
displacement x as a function of time t is
(5)
x(t) = h sin(ωt − π /2) .
€
1
Equation (4) or (5) describes an oscillatory motion for the block around its
equilibrium position x=0, with angular frequency of ω given by (3). The
period of this motion is
2π
m
T=
= 2π
.
(6)
ω
k
€
€
€
€
€
The velocity of the block is
dx(t)
v(t) =
= hω cos(ωt − π /2).
(7)
dt
The elastic potential energy, kinetic energy, and total mechanical energy are
given as
1
1
U = kx 2 (t) = kh 2 sin 2 (ωt − π /2) .
(8)
2
2
1
1
1
KE = mv2 (t) = m(hω )2 cos 2 (ωt − π /2) = kh 2 cos 2 (ωt − π /2) . (9)
2
2
2
1
1
1
E = U + KE = kh 2 sin 2 (ωt − π /2) + kh 2 cos 2 (ωt − π /2) = kh 2 . (10)
2
2
2
In equation (9), we substituted equation 3 for ω.
Equation (10) indicates that total mechanical energy is a constant (i.e., E is
conserved), while U and KE vary with time as in (8) and (9).
Remarks: The second order ODE in (2) or x˙˙ + ω 2 x = 0 always implies an
oscillatory motion for x with angular frequency ω. Notice that “+” sign in
this equation is the key to the oscillatory motion – which leads to sinusoidal
solutions.
€
Let’s consider pendulum motion. Is it harmonic oscillation with sinusoidal
motions?
2
Small oscillation pendulum motion.
For the pendulum below, suppose that θ is small and θ is positive when the
pendulum is at the right and negative when the pendulum is at the left.
Derive how θ varies with time.
€
€
€
€
The x component of the Newton’s 2nd law:
dv
−T sin θ = m x ,
(11)
dt
dθ
dθ
v x = lω cos θ = l cos θ
≈l ,
(12)
dt
dt
where in (12), we considered for small θ, cosθ =1.
Substitute (12) into (11) and also consider for small θ, sinθ = θ, and T=mg,
we have:
d 2θ
d 2θ g
,
or
(13)
−mgθ = ml
+ θ = 0,
dt 2
dt 2 l
Equation (13) takes the form of the 2nd order ODE for oscillatory motion.
Define ω2=g/l, and the solution for θ is
(14)
θ (t) = θ0 sin(ωt + φ0 ) .
€
The harmonic oscillation around the equilibrium position θ=0 is only
possible for pendulum motion of small amplitude.
v x (t) = lθ˙(t) = lθ0ω cos(ωt + φ0 ) .
€
€
(15)
1
1
U = mgl(1 − cos θ ) = mgl(1 − cos θ ) = mgl[1 − (1 − θ 2 + ...)] = mglθ 2 . (16)
2
2
2
In (16), we considered Taylor expansion of cos θ = 1-1/2θ …, near
θ=0. Therefore, the gravitational potential energy is
3
€
€
€
€
1
1
U = mglθ 2 = mglθ0 2 sin 2 (ωt + φ0 ) .
(17)
2
2
For small θ, the total velocity may be approximated by horizontal velocity
vx, and the kinetic energy, according to (15) for vx,
1
1
1
KE = mv2 = mv x 2 = mglθ0 2 cos 2 (ωt + φ0 ) ,
(18)
2
2
2
where we use the definition of ω2=g/l.
The total mechanical energy is
1
E = U + KE = mglθ0 2 ,
(19)
2
which is a constant, i.e., the total energy is conserved. However, potential
energy and kinetic energy exchange with each other all the time.
Physical pendulum.
An object with mass M is pivoted at a
distance l from the CM. The moment of
inertia around the CM is Icm. For small
amplitude oscillation, find its angular
frequency ω.
The torque produced by the gravity
around the pivot or rotation axis is
(20)
τ = −lMg sin θ ,
where the negative sign is due to the
clockwise rotation the torque would
produce (see figure). The moment of
inertia around the pivot point or
rotation axis is I=Icm+Ml2, according to the parallel axis theorem.
d 2θ
d 2θ
,
or −lMg sin θ = I
.
(21)
τ = Iα = I
2
2
dt
dt
For small oscillation, sin θ ≈ θ , and define ω 2 = lMg / I = lMg /(Icm + Ml 2 ) .
d 2θ
€
€
(22)
+ ω 2θ = 0 ,
€
dt
€ harmonic oscillation
which implies
€ with θ = θ0 sin(ωt) , and the angular
lMg
frequency is ω =
.
Icm + Ml 2
€
2
€
4