Foundation course - PHYSICS
Lecture 7-2: Harmonic motion
● simple harmonic motion
● velocity and acceleration of oscillations
● energy of oscillations
● pendulums
● damped and forced oscillations
Naďa Špačková spackova@physics.muni.cz
Simple harmonic motion
Simple harmonic motion
Frequency f:
● number of oscillations that
are completed each second
Period T:
● time for one complete oscillation (or cycle)
f =
1
T
Displacement x: x(t)= xm cos(ωt + ϕ)
Angular frequency ω:
ω =
2π
T
= 2π f
After one period T the displacement x
must return to its initial value:
x(t) = x(t + T) for all t
ω t + 2π =ω (t + T )
ω T = 2π
Simple harmonic motion
Amplitude xm
:
● maximum displacement
● positive constant whose value depends on how the motion was started
● displacement x varies between ± xm
(cosine varies between ± 1)
Phase (ωt + Φ):
● angular frequency ω
● phase constant (phase angle) Φ:
● depends on the displacement and
velocity of the particle at time t = 0
Simple harmonic motion
Blue curve: Φ = 0 in all three graphs
Difference in amplitude: x’m
> xm
Difference in period: T’ = T/2
Difference in phase
angle: Φ = -π/4
A particle undergoing simple harmonic oscillation of period T is at -xm
at time t = 0.
Is it at -xm
, at +xm
, at 0, between -xm
and 0, or between 0 and +xm
when
(a) t = 2.00T,
(b) t = 3.50T,
(c) t = 5.25T ?
Example:
Velocity
vx(t)=−ω xm sin(ωt + ϕ)
Velocity: vx (t) =
dx
dt
Displacement x:
x(t)= xm cos(ω t + ϕ)
Acceleration
vx(t) =−ω xm sin(ω t + ϕ)
ax (t)=
dvx
dt
Acceleration:
ax (t)=−ω2
xm cos(ωt + ϕ)
ax (t)=−ω2
x(t)
Acceleration is proportional to the
displacement but opposite in sign
Displacement x:
x(t)= xm cos(ω t + ϕ)
Velocity:
Velocity and acceleration
velocity amplitude
acceleration amplitude
The force law for harmonic motion
ax =−ω2
x
Fx = ma =−mω2
x
Fx =−k x
k = m ω2
ω =
√k
m
T = 2π
√m
k
Linear harmonic oscillator
Fx
is proportional to x rather than to some other power of x
Which of the following relationships between the force F on a particle and
the particle’s position x implies simple harmonic oscillation:
(a) F = -5x,
(b) F = -400x2
,
(c) F = 10x,
(d) F = 3x2
?
Example:
A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m.
The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless
surface and released from rest at t = 0.
(a) What are the angular frequency, the frequency, and the period of the resulting motion?
(b) What is the amplitude of the oscillation?
(c) What is the maximum speed vm
of the block and where is the block when it has this speed?
(d) What is the magnitude am
of the maximum acceleration of the block?
(e) What is the phase constant Φ for the motion?
(f) What is the displacement function x(t) for the spring – block system?
Example: simple harmonic oscillator
Energy of harmonic oscillator
The energy of a linear oscillator transfers back and forth between
kinetic energy Ek
and potential energy Ep
, while the sum of Ek
and
Ep
(= mechanical energy E of the oscillator) remains constant
Ep
Ep
Ek Ek
Ep
(t) + Ek
(t)Ep
(t) + Ek
(t) Ep
(x) + Ek
(x)
Potential energy is
entirely associated
with the spring
Kinetic energy is
entirely associated
with the block
Energy of harmonic oscillator
Potential energy Ep
: Ep =
1
2
k x
2
=
1
2
k xm
2
cos
2
(ωt + ϕ)
Kinetic energy Ek
: Ek =
1
2
m v
2
=
1
2
k xm
2
sin
2
(ωt + ϕ)
ω2
=
k
m
we used substitution
Mechanical energy E: E = Ep + Ek =
1
2
k xm
2
The mechanical energy of a linear oscillator is constant and
independent of time
In the figure, the block has a kinetic energy of 3 J and the spring has an elastic potential
energy of 2 J when the block is at x = +2.0 cm.
(a) What is the kinetic energy when the block is at x = 0?
(b) What is the elastic potential energy when the block is at x = -2.0 cm?
(c) What is the elastic potential energy when the block is at x = -xm
?
Example:
Example: kinetic and potential energy of simple harmonic oscillator
The block has mass m = 2.7 kg and oscillates at frequency f = 10.0 Hz and with
amplitude xm
= 20.0 cm.
(a) What is the total mechanical energy E of the spring – block system?
(b) What is the block’s speed as it passes through the equilibrium point?
Angular harmonic oscillator
Torsion pendulum Restoring torque
τ =−κ θ
torsion constant dependent on
the length, diameter, and material
of the wire
T = 2π √I
κ
Period of oscillations:
moment of inertia
Simple pendulum
Simple pendulum:
● consists of a particle of mass m (called the bob of the pendulum)
suspended from one end of an unstretchable, massless string of length L
that is fixed at the other end
Two acting forces:
● gravitational force
● tension from the string
This component produces
a restoring torque about
a pendulum pivot
Simple pendulum
τ =−L(Fg sinθ )
Restoring torque:
I α =−L(m gsinθ )
I…moment of inertia, α...angular acceleration
We assume the angle θ is small: then sin θ ≈ θ
α =−
m g L
I
θ
ax(t)=−ω2
x(t)It is similar to the equation:
Simple pendulum swinging through only small
angles is approximately simple harmonic
oscillator
T = 2π
√ I
m g L
All mass of the simple pendulum
is concentrated in the bob:
I = m L2
T = 2π
√L
g
Period of the simple pendulum
swinging is not dependent on the
mass of the bob
Period of swinging:
Physical pendulum
T = 2π
√ I
m g L
Period of swinging of the physical pendulum:
I is not simply mL2
because
it depends on the shape of
the physical pendulum, but
it is proportional to m
The physical pendulum does not swing
if it pivots at its center of mass
The physical pendulum that oscillates about a
given point O with period T is a simple pendulum
of length L0
with the same period T – the point at
distance L0
is called the center of oscillation
Three physical pendulums, of masses m0
, 2m0
, and 3m0
, have the same shape
and size and are suspended at the same point. Rank the masses according to
the periods of the pendulums, greatest first.
Example:
Simple harmonic motion and uniform circular motion
Simple harmonic motion is the projection of uniform circular motion
on a diameter of the circle in which the circular motion occurs.
x(t) = xm cos(ωt + ϕ) vx (t) =−ω xm sin(ωt + ϕ) ax (t) =−ω2
xm cos(ωt + ϕ)
Damped oscillations
Mechanical energy of the block – spring
system decreases: energy is
transferred to thermal energy of the
liquid and vane
Damping force: Fd =−bv
b...damping constant, v...velocity
The force on the block from the spring: Fs =−k x
ma =−bv−k x
underdamped
critically damped
overdamped osc.
ω =
√ω0 −(
b
2m
)
2
Angular frequency of a damped oscillator:
Forced oscillations and resonance
● free oscillations
● natural angular frequency ω causing the free oscillation
● driven (forced) oscillations
● angular frequency ωd
of the external driving force causing the
driven oscillations
● the system oscillates with angular frequency ωd
● the velocity amplitude is greatest when ωd
= ω
● the amplitude xm
of the system is
(approximately) greatest under the same
condition.
...this condition is called resonanceωd = ω