1. Physics 231
Topic 7: Oscillations
Alex Brown
October

2. Key Concepts: Springs and Oscillations
Periodic Motion
Frequency & Period
Simple Harmonic Motion (SHM)
Amplitude & Period
Angular Frequency
SHM Concepts
Energy & SHM
Uniform Circular Motion
Simple Pendulum
Damped & Driven Oscillations Covers chapter 7 in Rex & Wolfson

3. Springs
xo=0

4. Hooke’s Law xo=0 Fs = -kx Hooke’s law k = spring constant in units
of N/m.
x = displacement from
the equilibrium point xo=0

5. How to find the spring constant k
When the object hanging from the spring is not moving:
∑F = ma = 0
Fspring + Fgravity = 0
-k(d) + mg = 0
k = mg/d x
x = 0
Fspring = -kx
x = d
Fgravity = mg

6. W = area under F vs x plot = ½kx2
Springs
Hooke’s Law:
Fs = -kx
k: spring constant (N/m)
x: distance the spring is stretched from equilibrium
W = area under F vs x plot = ½kx2
The energy stored in a spring depends on how far the spring has been stretched: elastic potential energy
PEspring = ½ k x2

7. PINBALL! PEspring = ½ k x2 PEgravity = mgh KE = ½ mv2 ME = PE + KE
The ball-launcher spring has a
constant k = 120 N/m.
A player pulls the handle 0.05 m.
The mass of the ball is 0.1 kg.
What is the launching speed?
PEspring = ½ k x2
PEgravity = mgh
KE = ½ mv2
ME = PE + KE
= Constant
(PEgravity+PEspring+KEball)pull = (PEgravity+PEspring+KEball)launch
mghpull + ½kxpull2 + ½mvpull2 = mghlaunch + ½kxlaunch2 + ½mvlaunch2
½120(0.05)2 = 0.1 (9.81) ( 0.05*sin(10o) ) + ½(0.1)vlaunch2
0.15 = 8.5x v2
v =1.7 m/s

8. Conservation of momentum (closed system) p1i + p2i = p1f + p2f
Momentum p = mv
Impulse p = pf - pi = Ft
Conservation of momentum (closed system) p1i + p2i = p1f + p2f
Collisions in one dimension given
given m1 and m2 or c = m2 /m1
v1i and v2i
Inelastic (stick together, KE lost)
vf = [v1i + c v2i] / (1+c)
Elastic (KE conserved)
v1f = [(1-c) v1i + 2c v2i] / (1+c)
v2f = [(c-1) v2i + 2v1i] / (1+c)

9. Collisions in one dimension given given m1 = m2 or c = 1 v1i and v2i
Inelastic (stick together, KE lost)
vf = [v1i + v2i] / 2
Elastic (KE conserved)
v1f = v2i
v2f = v1i

10. Carts on a spring track k = 50 N/m v = 5.0 m/s m1 = m2 = 0.25 kg v m1
|||||||||||
What is the maximum compression of the spring
if the carts collide a) elastically and b) perfectly inelastic?
Conservation of momentum and KE with c = 1
v1f = v v2f = :: v1f = v2f = v = 5 m/s
Conservation of energy: ½mv2 = ½kx2
(0.5) (0.25) (5)2 = (0.5)(50)x x = 0.35 m
B) Conservation of momentum only with c = 1
vf = (v1f + v2f)/2 = (v + 0)/2= 2.5 m/s
Conservation of energy: ½mvf2 = ½kx2
(0.5)(0.5)(2.5)2 = (0.5)(50)x x=0.25 m

11. Hooke’s Law Fs = -kx Hooke’s law
If there is no friction, the mass continues to oscillate back and forth.
If a force is proportional to the displacement x, but
opposite in direction, the resulting motion of the object is
called: simple harmonic oscillation

12. Simple harmonic motion
displacement x A c) b)
time (s) a)
Amplitude (A): maximum distance from
equilibrium (unit: m)
Period (T): Time to complete one full
oscillation (unit: s)
Frequency (f): Number of completed
oscillations per second. f=1/T
(unit: 1/s = 1 Herz [Hz])
Angular frequency  = 2πf = 2π/T
(c)
(b)
(a)

13. Concept Quiz displacement x 5 m 2 4 6 8 10 time (s) -5 m
what is the amplitude of the harmonic oscillation?
what is the period of the harmonic oscillation?
what is the frequency of the harmonic oscillation?
Amplitude: 5 m
period: time to complete one full oscillation: T = 4s
frequency: number of oscillations per second
f = 1/T = 0.25/s = 0.25 Hz

14. What the equation? x = A cos(t) A = 5 T = 2π
displacement x
5 m 2 4 6 8 10
time (s)
-5 m
x = A cos(t)
A = T = 2π
 = 2π/T = 2π/4 = 1.57 rad/s

15. Clicker Quiz!
A mass on a spring in SHM has amplitude A and period T. What is the total distance traveled after a time interval of 2T?
A/2 A 4A 8A
In one cycle covering period T, the mass moves from +A to -A and back: 2A+2A = 4A
In two periods, the mass moves twice this distance: 8A

16. Energy and Velocity Ekin(½mv2) Epot,spring(½kx2) Sum 0 ½kA2 ½kA2 x=+A
½mv2max ½mv2max
½k(-A) ½kA2
x=+A
x=0
x=-A
conservation of ME: ½mv2max = ½kA so vmax = ±A(k/m)

17. Velocity and acceleration in general
Total ME at any displacement x: ½mv2 + ½kx2
Total ME at max. displacement A: ½kA2
Conservation of ME: ½kA2 = ½mv2 + ½kx2
So: v = ±[(A2-x2)k/m] also F = ma = -kx so a = -kx/m
position x
velocity v
Acceleration a +A
-kA/m
±A(k/m) -A
kA/m

18. An Example
The maximum velocity and acceleration of mass connected to a spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively.
Find the oscillation amplitude.

19. An Example
The maximum velocity and acceleration of mass connected to a spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively.
Find the oscillation amplitude.
We do not know the mass of the object or the spring constant. So we must find a way to eliminate these variables!
so vmax = ±A(k/m)
so (vmax )2 = A2 (k/m)
so amax = A (k/m)
A = vmax2/amax = (0.95)2/(1.56) = 0.58 m

20. Harmonic oscillations vs circular motion
x(t) = A cos 
r=A
The projection of the position of the circulating object on the x-axis as a function of time is the same as the position of the oscillating spring.

21. Harmonic oscillations vs circular motion
vx(t) = -v sin v  vx 
r=A
The vx projection of the linear velocity of the rotating object is the velocity of the mass on the spring.

22. x=0
x=-A
x=+A
But remember for circular motion
with constant speed around the
circle  = t and v = r = A
where  is the “angular frequency” 
So x(t) = A cos = A cos(t)
vx(t) = -v sin = - A sin(t)
time to complete one circle
= T = one period
so T = 2 or  = 2/T
r=A

23. Harmonic oscillations vs circular motion
The simple harmonic motion can be described by the
projection of circular motion on the horizontal axis.
x(t) = A cos(t)
v(t) = -A sin(t)
A amplitude of the oscillation
 = 2/T = 2f angular frequency
T period
f = 1/T frequency.

24. The angular frequency does not depend on A
x = A cos(t) A x
time (s) -A
velocity v = -A  sin(t)
A  v
time (s)
-A 
Remember thus
The angular frequency does not depend on A

25. Displacement vs Acceleration
displacement x A
time (s) -A
Newton’s second law: F=ma  -kx=ma  a=-kx/m
acceleration(a)
kA/m
-kA/m

26. v a x = A cos(t) A x time (s) -A v = -A  sin(t) time (s)
acceleration a = -A 2 cos(t)
A 2
time (s) a
-A 2

27. Period of a Spring
Increase m:
increase T
Increase k:
decrease T

28. Clicker Quiz! C
A mass is oscillating horizontally while attached to a spring
with spring constant k. Which of the following is true?
When the magnitude of the displacement is largest,
the magnitude of the acceleration is also largest.
b) When the displacement is positive, the acceleration is
also positive
c) When the displacement is zero, the acceleration is
non-zero

29. The pendulum
x=-A
x=+A
x=0

30. Conservation of Energy! v2 = 2gh velocity at the bottom
The pendulum
 = max = A
y=h
Conservation of Energy!
At x=+A: PE = mgh KE = ½mv2 = 0
At x=0: PE = mg(0) = 0 KE = ½mv2
½mv2 = mgh
v2 = 2gh velocity at the bottom

31. The pendulum Tension at max angle  = max T = mg cosmax
Tension at bottom  = 0
T = mg + mv2/L L   T s
mg sin 
mg cos

32. The pendulum Restoring force: F=-mg sin The force pushes the mass m
back to the central position. L  T s
mg sin 
mg cos

33. The pendulum Restoring force: F=-mg sin The force pushes the mass m
back to the central position.
sin   (radians) if  is small
F = - mg also s = L 
so: F=-(mg/L)s L  T s
mg sin 
mg cos

34. pendulum vs spring The pendulum Does not depend on mass
spring pendulum

35. Damped Oscillations
Newton’s 1st Law: An object will stay in motion unless acted upon by a force.
SHM is the same: oscillations will last forever in the absence of additional forces.
Common force of friction often “damps” oscillations and brings them to a stop.
In homework A is reduced by some factor f after one period
After one oscillation A1 = f A
So for n oscillations An = fn A
mechanical energy ME = (1/2) k A2

36. Driven Oscillations The same can be applied in reverse:
An object NOT in motion can be put into oscillation by applying a force
This is known as driven oscillations.
If the force acts with the same frequency as the as that of the object this is known as resonance.
With driven oscillation, the amplitude grows!

37. Spring problem
A bungee jumper with height h=2 m and mass m = 80 kg leaps from a bridge with height H = 30 m. A bungee cord with spring constant k = 100 N/m attached to his legs. What is the maximum length the cord needs to be if he is to avoid hitting the water below?
Define A = extended “amplitude” of the bungee cord
Total extension = jumper’s height + nominal length of the cord
+ extended length of the cord = h+L+A
Must stop before h+L+A = H = 30m, or A = H-L-h
Use conservation of ME: ½mv2 + ½kx2 + mgH
On top of the bridge: ME = mgH
Maximum extension of the cord:
ME = ½ kA2 = ½ k(H-L-h)2
ME(bridge) = ME(ground)
mgH = ½k(H-L-h) 2
(80 kg)(9.81 m/s2)(30m) = ½(100 N/m)(30m-L-2m) 2
= 50 (28m-L) 2
(28m-L) = sqrt(23544/50) = 21.7 m L = 6.3 m

38. Spring problem
A h=2m tall, m=80 kg bungee jumper leaps from a H=30m bridge with a bungee cord with spring constant k = 100 N/m attached to his legs.
What is the frequency of his subsequent SHM?
𝐓 𝐬𝐩𝐫𝐢𝐧𝐠 =𝟐𝛑 𝐦 𝐤
= 𝟐𝛑 𝟖𝟎 𝟏𝟎𝟎 = 5.62 sec
Frequency f = 1/T
= 1/5.62s = 0.18 Hz

39. Spring problem x(t) = Acos(t) = 0.1 cos(22.4t)
A mass of 0.2 kg is attached to a spring with k=100 N/m.
The spring is stretched over 0.1 m and released.
What is the angular frequency () of the corresponding
motion?
What is the period (T) of the harmonic motion?
What is the frequency (f)?
What are the functions for x,v and t of the mass
as a function of time? Make a sketch of these.
 =(k/m) = (100/0.2) = 22.4 rad/s
 = 2/T T= 2/ = 0.28 s
 = 2f f=/2 = 3.55 Hz (=1/T)
x(t) = Acos(t) = cos(22.4t)
v(t) = -Asin(t) = sin(22.4t)
a(t) = -2Acos(t) = cos(22.4t)

40. 0.1 x
0.28
0.56
time (s)
-0.1
velocity v
2.24
0.28
0.56
-2.24
50.2 a
0.28
0.56
-50.2

41. pendulum problem The machinery in a pendulum clock is kept
in motion by the swinging pendulum.
Does the clock run faster, at the same speed,
or slower if:
The mass is hung higher
The mass is replaced by a heavier mass
The clock is brought to the moon
The clock is put in an upward accelerating
elevator? L m
Moon
Elevator
Faster(A)
Same(B)
Slower(C)

42. Pendulum Quiz
A pendulum with length L of 1 meter and a mass of 1 kg, is oscillating harmonically. The period of oscillation is T.
If L is increased to 4 m, and the mass replaced by one of 0.5 kg, the period becomes:
A) 0.5 T
B) 1 T
C) 2 T
D) 4 T
E) 8 T
Mass does not matter

43. Spring problem
A mass of 1 kg is hung from a spring. The spring stretches
by 0.5 m. Next, the spring is placed horizontally and fixed
on one side to the wall. The same mass is attached and the
spring stretched by 0.2 m and then released. What is
the acceleration upon release?
1st step: find the spring constant k
Fspring =-Fgravity or -kd =-mg
k = mg/d = (1)(9.8)/0.5 = 19.6 N/m
2nd step: find the acceleration upon release
Newton’s second law: F=ma  -kx = ma  a = -kx/m
a = -(19.6)(0.2)/1 = m/s2

44. Sprng problem x d A block with mass of 200 g is placed
over an opening. A spring is placed under the opening and compressed by a distance d=5 cm. Its spring constant is 250 N/m. The spring is released and launches the block. How high will it go relative to its rest position (h)? (the spring can be assumed massless) d
ME = ½mv ½kd mgx must be conserved.
Initial: v = d = m x = m = 0.2 kg
(0.5)(250)(0.05) =
Final: v = d = x = h
(0.2)(9.8)(h) = h
Conservation of ME: = 1.96 h so h = 0.16 m