Physics 231 Lecture 17
•  Main points of today’s lecture:
•  Springs and masses
•  Simple harmonic motion of a spring:
•  Pendulum
T = 2π
L
g
θ = θ max cos(2πft + ϑ0 )
x = A cos(ωt + θ 0 )
vx = −ωA sin(ωt + θ 0 )
a x = −ω 2 A cos(ωt + θ 0 )
ω 2 = k / m θ 0 and A are constants
Concept Quiz
displacementx
5m
2
4
6
8
-5m
whatistheamplitudeoftheharmonicoscilla!on?
a) 
b) 
c) 
d) 
5m
10m
2s
4s
10!me(s)
Concept Quiz
displacementx
5m
2
4
6
8
-5m
whatistheperiodoftheharmonicoscilla!on?
a) 
b) 
c) 
d) 
4s
2s
.25Hz
.5Hz
10!me(s)
Concept Quiz
displacementx
5m
2
4
6
8
-5m
whatisthefrequencyoftheharmonicoscilla!on?
a) 
b) 
c) 
d) 
f=1/T=0.25/s=0.25Hz
f=1/T=0.5/s=0.5Hz
f=2π/T=π/2/s=1.57Hz
f=2π/T=0.25π/s=0.78Hz
10!me(s)
Frequency vs Period
A
-A
time (s)
FrequencyvsPeriod
Timetogofrom+Ato–Aandbackistheperiod.i.e.theperiodisthe
!meittakestocompleteonefullcycle.Period=T
Frequency:f=1/T.eg,ifT=2secèf=½Hz
Hz=1/s
Iffislarge,periodissmall: f=1kHz=1000cyclespersecond
T=1/(1000Hz)=0.001sec Period and Frequency
•  Sinusoidal functions are periodic
–  Adding 2π to the argument does not change the value of the function
A
-A
• 
A cos (θ ) = A cos ( 2π + θ )
θ (rad)
θ = ωt
⎛ ⎛ 2π
⎞⎞
A cos (ωt ) = A cos ( 2π + ωt ) = A cos ⎜ ω ⎜
+ t ⎟⎟
⎠⎠
⎝ ⎝ω
The period is the time interval over which function repeats itself.
2π
T=
ω
Question 7.1 Harmonic Motion I
A mass on a spring in SHM has
a) 0
amplitude A and period T. What is
b) A/2
the total distance traveled by the mass
c) A
after a time interval T?
d) 2A
e) 4A
Question 7.1 Harmonic Motion I
A mass on a spring in SHM has
a) 0
amplitude A and period T. What is
b) A/2
the total distance traveled by the mass
c) A
after a time interval T?
d) 2A
e) 4A
In the time interval T (the period), the mass goes
through one complete oscillation back to the starting
point. The distance it covers is A + A + A + A (4A).
Graphical Representation of Motion
•  Our equations:
•  x=A cos(θ)=A cos(ωt+ θ0)
What is θ0?
x
• 
• 
• 
• 
a) 0 rad
cos
b) π/2 rad
c) π rad
d) 3π/2 rad
(ω ⎡⎣3T / 4 ⎤⎦ + θ ) = 1
0
ω ⎡⎣3 / 4 ⎤⎦ T + θ 0 = ±2π n
ω ⎡⎣3 / 4 ⎤⎦ ⎡⎣ 2π / ω ⎤⎦ + θ 0 = ±2π n
θ 0 = ±2π n − 3π / 2 = π / 2 ± 2π m
If θ0 the right time dependence then θ0=±2πn also will as well
Recall ωT=2π
Example
The motion of an object is described by the equation
x = (0.30 m) cos(πt/3),
where t is assumed to be in seconds. Find (a) the position, (b)
velocity and (c) acceleration of the object at t = 0 and t = 0.60 s,
(d) the amplitude of the motion, (e) the frequency of the motion,
and (f) the period of the motion.
general form: x = A cos(ω t + θ0 ); A = 0.3m; θ0 = 0
ω = π / 3Hz = 1.05Hz, f = ω / (2π ) = .17Hz; T = (1/ f ) = 6s
x = A sin(ω t + θ0 ) = ( 0.3m ) sin(1.05t) (t is in sec.)
v = -ω Asin(ω t + θ0 ) = − (1.05Hz )( 0.3m ) sin(1.05t)
a = −ω2 x = −1.1x
at t = 0 : x = 0.3 m; v = 0 ; a = −0.33m / s 2
at t = 0.6s : x = 0.24 m; v = −0.37m / s ; a = −.26m / s 2
Graphical Representation of Motion
•  x=A cos(θ)=A cos(ωt+ θ0)
•  vx= -ωAsin (ωt+ θ0)
•  ax= - ω2Acos(ωt+ θ0)= - ω2x
•  When x is a maximum or minimum,
velocity is zero
•  When x is zero, the speed is a
maximum
•  When x is a maximum in the
positive direction, a is a maximum
in the negative direction
Conceptual question
•  A mass attached to a spring oscillates back and forth as
indicated in the position vs. time plot below. At point P, the
mass has
– 
– 
– 
– 
– 
a. positive velocity and positive acceleration.
b. positive velocity and negative acceleration.
c. positive velocity and zero acceleration.
d. negative velocity and positive acceleration.
e. negative velocity and negative acceleration.
recall
ax = −k / m i x
Example
•  A 50 coil spring has a spring constant of 860 N/m. One end of a 50coil spring is attached to a wall. An object of mass 45 kg is attached to
the other end of the spring and the system is set in horizontal
oscillation. What is the angular frequency of the motion?
–  a) 2.39 Hz
–  b) 4.37 Hz
860N / m
k
=
= 4.37Hz
–  c) 5.21 Hz
ω=
45kg
m
–  d) 6.85 Hz
–  e) 9.22 Hz
Reading quiz
•  If mass spring system was arranged vertically with the mass suspended
from the 50 coil spring, how would the frequency change?
–  a) it would be smaller because gravity subtracts from the spring
force at the bottom of its motion.
–  b) it would be larger because gravity adds to the spring force at the
top of its motion.
–  c) it would be exactly the same. Gravity only displaces the
equilibrium point so that the equilibrium length is greater.