Chapter 11: Vibrations & Waves
• First half of Chapter: Vibrations
• Second half: Waves
• Chapter 12: Sound waves.
Sect. 11-1: Simple Harmonic Motion
• Vibration  Oscillation = back & forth motion
of an object
• Periodic motion: Vibration or Oscillation that
regularly repeats itself.
• Simplest form of periodic motion: Mass m
attached to ideal spring, spring constant k
(Hooke’s “Law”: F = -kx) moving in one
dimension (x).
– Contains most features of more complicated
systems  Use the mass-spring system as a
prototype for periodic oscillating systems.
Mass-Spring System
• Mass-spring (no friction):
Simplest form of periodic
motion. Prototype for ALL
vibrating systems
• Restoring force obeys
Hooke’s “Law”: F = -kx
• A “simple harmonic
oscillator” (SHO)
• Undergoing “simple
harmonic motion” (SHM)
• Spring constant k: Depends
on spring.
Terminology & Notation
• Displacement = x(t)
• Velocity = v(t)
• Acceleration = a(t)  Constant!!!!!!!
 Constant acceleration equations from Ch. 2
DO NOT APPLY!!!!!!!!!!!
•
•
•
•
Amplitude = A (= maximum displacement)
One Cycle = One complete round trip
Period = T = Time for one round trip
Frequency = f = 1/T = # Cycles per second
Measured in Hertz (Hz)
A and T are independent!!!
• Repeat: Mass-spring system contains most
features of more complicated systems
 Use mass-spring system as a prototype for
periodic oscillating systems.
• Results we get are valid for many systems
besides this prototype system.
• ANY vibrating system with restoring force
proportional to displacement (F = -kx) will
exhibit simple harmonic motion (SHM) and
thus is a simple harmonic oscillator (SHO).
(whether or not it is a mass-spring system!)
Mass-Spring System
Vertical Mass-Spring System
Sect. 11-2: Energy in the SHO
• A Ch. 6 result: Compressing or stretching
an ideal spring a distance x gives elastic PE:
PE = (½)kx2
 Total mechanical energy of SHO (massspring system) is conserved (a constant!)
E = KE + PE = (½)mv2 + (½)kx2 = const
– Assuming that friction is neglected!
Energy conservation
E = (½)mv2 + (½)kx2 = constant
– The same throughout the motion
• Mass is moving back & forth.
 Clearly x = x(t)
The mass is speeding up & slowing down.
 Clearly v = v(t)
 E = (½)m[v(t)]2 + (½)k[x(t)]2
But E = const  E(t) (independent of t!)
E = (½)mv2 + (½)kx2 = constant
• Conservation of KE + PE gives some properties:
• At maximum x: At x = A, v = 0.

Ex=A = (½)k(A)2
• At maximum v: At x = 0, v = v0

Ex=0 = (½)m(v0)2
• Somewhere in between (any x: -A  x  A).
Ex = (½)mv2 + (½)kx2
• But E = constant

Ex=A = Ex=0 = Ex = E
E = (½)mv2 + (½)kx2 = constant
• Using results that Ex=A = Ex=0 = Ex = E :
(½)k(A)2 = (½)m(v0)2 = (½)mv2 + (½) kx2
• Relation between maximum x (x=A) &
maximum v (v = v0): (½)k(A)2 = (½)m(v0)2 
v0 =  (k/m)½A
• Velocity at any x: (½)k(A)2 = (½)mv2 + (½)kx2

v =  v0 [1 - (x2/A2)]½
• Position at any v: (½)m(v0)2 = (½)mv2 + (½)kx2
 (using above for v0): x =  A[1 - (v2/v02)]½
Energy in Mass-Spring System
Examples 11-4 & 11-5
Everything holds also for a vertical mass-spring system!
Sect. 11-3: Period & Nature of SHM
• Experiment: Period T of SHO does NOT
depend on amplitude A!
– T depends only on m & k
• Derivation: Use motion in a circle &
compare SHM to object rotating in circle.
• Will give us T and expression for x(t).
Circular Motion Comparison
Similar triangles  (v/v0) = [A2 - x2]½/(A)
or
v = v0 [1 - (x2/A2)]½
Same as for SHM!
 Projection on x-axis of motion of mass moving
in a circle is the same as for a mass on a spring!
• Period T for one revolution of circle  Period
T for SHO!
• Circle, radius A, circumference 2πA, velocity
v0. Mass goes once around (distance 2πA) in
time T  2πA = v0T. Or,
T = (2πA)/(v0)
(1)
• From energy discussion: (½)k(A)2 = (½)m(v0)2

v0 = (k/m)½ A
(2)
• Combining (1) & (2) gives:
T = 2π(m/k)½  Period of SHO
Period of SHO:
T = 2π(m/k)½
– Agrees with experiment!
– Independent of amplitude A
Frequency f = (1/T)
f = [1/(2π)](k/m)½
Angular frequency: ω = 2πf
ω = (k/m)½
Sect. 11-3: x(t)
• Find x(t)
• From diagram:
x = A cos(θ)
• Mass rotating with
angular velocity ω

θ = ωt, ω = 2πf
 For SHO:
x = Acos(ωt) = Acos(2πft) = Acos[(2πt)/(T)]
• At t = 0, x = A. At t = T, x = A
-A  x  A since -1  cos(θ)  1
x = Acos(ωt) = Acos(2πft) = Acos[(2πt)/(T)]
• Assumes initial conditions that x = A at t = 0. Can
show, if x is something else at t = 0, x(t) is still a
sine or cosine function: If, at t = 0, x = 0, find:
x = Asin(ωt) = Asin(2πft) = Asin[(2πt)/(T)]
x(t), v(t), a(t)
• For x = A at t = 0:
x(t) = Acos(2πft) = Acos[(2πt)/(T)]
– x is clearly a function of time x = x (t)!
– Velocity and acceleration are ALSO functions of time:
v = v(t) , a = a(t). Also sinusoidal functions.
• v(t): We had v = v0 [1 - (x2/A2)]½. Put x(t) in this
to get v(t):
[1 - (x2/A2)] = 1 - cos2 [(2πt)/(T)] = sin2 [(2πt)/(T)]
(Trig identity!)

v(t) = v0 sin[(2πt)/(T)]
Only for x = A at t = 0!
• For x = A at t = 0:
x(t) = Acos(2πft) = Acos[(2πt)/(T)]
• a(t): We had (Hooke’s “Law”):
F = -kx = -kx(t)
We also have Newton’s 2nd Law:
F = ma
Combining gives a = a(t) = -(k/m) x(t)
Using above form for x(t):
a(t) = -(k/m) Acos[(2πt)/(T)]
NOTE! The acceleration is a function of time
and is NOT constant! a  constant !
• For x = A at t = 0:
a(t) = -(k/m) Acos[(2πt)/(T)]
a = a(t)  constant !
• NOTE!! This means that the 1 dimensional
kinematic equations for constant acceleration
(from Chapter 2) DO NOT APPLY!!!!!!
• That is, THROW THESE AWAY FOR
THIS CHAPTER!!
THESE ARE WRONG AND WILL GIVE
YOU WRONG ANSWERS!!
Summary
• We’ve shown that for x = A at t = 0:
x(t) = Acos[(2πt)/(T)], v(t) = Asin[(2πt)/(T)]
a(t) = -(k/m) Acos[(2πt)/(T)]
• From conservation of energy: (½)k(A)2 = (½) m(v0)2
• Here:
x(t) = Asin[(2πt)/(T)]
v(t) = Acos[(2πt)/(T)]
a(t)
= -(k/m) Asin[(2πt)/(T)]
x(t), v(t), a(t) :
Functional forms depend
on initial conditions!