Question
A “mass on a spring” problem consists of a mass m attached to a spring with a spring constant k .
If the mass m increases by a factor of 2, the angular frequency of oscillation of the mass __.
1. is doubled
2. is multiplied by a factor of 2 1/2
3. is halved
4. is multiplied by a factor of 1/2 1/2

Question
A “mass on a spring” problem consists of a 2.0 kg mass attached to a spring with a spring constant of 200 N/m.
The angular frequency of oscillation of the mass is __ rad/s.
1.
2
2.
10
3.
60
4. 100

Question
An oscillating mass has a position x versus time t given by x = (0.5 m) cos[(3.1 rad/s) t + 0.1 rad]
The period of the oscillation is __ s.
1.
1.0
2.
1.5
3.
2.0
4.
3.1

A 0.20 kg mass undergoes simple harmonic motion along the x -axis as shown.
Question
The maximum speed of the mass as time passes is __ m/s.
1.
1.5
2.
3.1
3.
6.3
4. 12.6

A 0.20 kg mass undergoes simple harmonic motion along the x -axis as shown.
T = 2.0 s

=
 rad/s
Question
The maximum magnitude of the acceleration of the mass as time passes is __ m/s 2 .
1.
3.8
2.
5.6
3.
7.3
4.
9.9

A 0.20 kg mass undergoes simple harmonic motion along the x -axis as shown.
T = 2.0 s

=
 rad/s
Question
The kinetic energy of the mass is maximum at time t = __ s.
1.
0.00
2.
0.25
3.
0.50
4.
0.75

A 0.20 kg mass undergoes simple harmonic motion along the x -axis as shown.
T = 2.0 s

=
 rad/s
Question
The potential energy of the system is maximum at time t = __ s.
1.
0.00
2.
0.25
3.
0.50
4.
0.75

A 0.20 kg mass undergoes simple harmonic motion along the x -axis as shown.
T = 2.0 s

=
 rad/s
Question
The mechanical energy of the oscillator is ___ J.
1.
1.0
2.
3.8
3.
5.6
4.
8.2

L mg

  x
L
F
  mg sin

F
  mg x
L
 ma d
2 x dt a
  g
L x
2
  g
L x
If
 
1 than x ( t )

  sin
  
A cos
  t
   g
L
; T

2

L g
L

0 .
5 m

0 v
0
Example:

10


0


0
L
0

 v
0
 g

9 .
8 m / s
2
L 0 .
5 m x


A
 
A cos sin




0
0







4 .
4 m / s

A
 
0
L

2


0 x ( t )


36 m
10

360

0 .
5 m
 cos
 
4 .
4 m / s
 t
  

36 m

Example: A person swings on a swing. When the person sits still, the swing moves back and forth at its natural frequency. If, instead, the person stands on the swing, the new natural frequency of the swing is:
A. Greater B. The same C. Smaller
  g
If the person stands, L becomes smaller .
L
Example: Grandpa decides to move to the Moon, and he naturally takes his old pendulum clock with him. But gravity on the Moon is approximately g/6... As a result, his clock is:
A. Too fast B. Too slow
C. Too fast until Noon, too slow after noon
T

2



2

L g
The period of the pendulum is longer on the Moon.
So each “second according to this clock” is then longer than a real second.

L

  x
L mg
I
 mL
2
L

  x
L mg


FL sin
   mgL sin
   mgL

 
I

   mgL

I
   max cos
  t
   d
2
 dt
2
  mgL

I
  mgL
T

I
2

I mgL
  g
L
T

2

L g
(a rigid body that oscillates about an axis)
L - distance between CM and suspension point
I - moment of inertia about suspension point
I

I
CM
 mL
2

Example: A rod of length 6L long has a disk with r = L attached to it.
The disk can be positioned along the rod’s length. Both the rod and the disk have mass m. The distance between the center of the disk and the pivot point is 3L in case A and 5L in case B. Find the ratio ω
A
/ ω
B.
  mgd
I
3L pivot pivot
A 5L


A
B

I
A
(2 )
B
I
B
 A B d I
A
r=L
6L
A:
B: d
A
 3 I
A d
B

2
 m
3
 4 I
B
Case A Case B


1
3
1
3 m L 2 m L 2


1
2
1
2 mL 2  m L 2  21.5
mL 2 mL 2  m L 2  37.5
mL 2

A

B

A

3 L  37.5
mL 2
4 L  21.5
mL 2
 

T
A
T
B



B
A

1
1.14
 0.874
(Ok, maybe not this one…)