If the mass m increases by a factor of 2, the angular frequency of oscillation of the mass __.

Question

1. is doubled

2. is multiplied by a factor of 21/2

3. is halved

4. is multiplied by a factor of 1/21/2

A “mass on a spring” problem consists of a mass m attached to a spring with a spring constant k.

The angular frequency of oscillation of the mass is __ rad/s.

Question

1. 2

2. 10

3. 60

4. 100

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 period of the oscillation is __ s.

Question

1. 1.0

2. 1.5

3. 2.0

4. 3.1

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 maximum speed of the mass as time passes is __ m/s.

Question

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.

The maximum magnitude of the acceleration of the mass as time passes is __ m/s2.

Question

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

The kinetic energy of the mass is maximum at time t = __ s.

Question

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

The potential energy of the system is maximum at time t = __ s.

Question

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

The mechanical energy of the oscillator is ___ J.

Question

1. 1.0

2. 3.8

3. 5.6

4. 8.2

A 0.20 kg mass undergoes simple harmonic motion along the x-axis as shown.

T = 2.0 s = rad/s

The simple pendulum

sinmgF If 1 than sin

maL

xmgF

xL

ga tAtx cos)(

g

LT

L

g 2 ;

mg

L

L

x

Example:

0

10

5.0

0

0

v

mL

smm

sm

L

g/4.4

5.0

/8.9 2

0sin0

0cos

0

0

Av

AxL

036

5.0360

1020

mmLA

xL

g

dt

xd

2

2

tsmmtx /4.4cos

36)(

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

L

g If the person stands, L becomes smaller.

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 slowC. Too fast until Noon, too slow after noon

The period of the pendulum is longer on the Moon. So each “second according to this clock” is then longer than a real second.

g

LT

22

The simple pendulum (method 2)

mg

L

L

x

I

mgLmgLFL

sinsin

I

mgL

dt

d

I

mgL

2

2

tcosmax

mgL

IT

I

mgL 2

g

LT

L

g 2

2mLI

mg

L

L

x

The physical pendulum(a rigid body that oscillates about an axis)

2mLII CM

L - distance between CM and suspension pointI - moment of inertia about suspension point

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 ωA/ωB.

pivot pivot

3L5L

6L

r=L

Case A Case B

mgd

I

A

A A BA

AB BB

B

(2 )

(2 )

m gdI d I

d Im gdI

2 2 2 2A A

1 13 (6 ) (3 ) 21.5

3 2d L I m L mL m L mL A:

2 2 2 2B B

5 3 1 14 (6 ) (5 ) 37.5

2 3 2m L m L

d L I m L mL m L mLm

B:

2A BA

2AB B

3 37.5 112.51.14

4 21.5 86

d I L mLd I L mL

BA

AB

10.874

1.14TT

Lowering the disk increases the period (ie, slows the pendulum down).

This is how you tune a grandfather clock.

(Ok, maybe not this one…)