Physics 42200

Waves & Oscillations

Spring 2013 SemesterMatthew Jones

Lecture 14 – French, Chapter 6

Vibrations of Continuous Systems

• Equations of motion for masses in the middle:

���� + 2��� − � ��� + ���� = 0��� + 2 �� ��� − �� � ��� + ���� = 0

• Proposed solution:

�� � = �� cos����� + ����

�� = −�� + 2 �� �

�� �• We solved this to determine �� and ��…

� � � �

• Amplitude of mass � for normal mode �:

��,� = � sin��� + 1

• Frequency of normal mode �:

�� = 2�� sin��

2 + 1• Solution for normal modes:

�� � = ��,� cos���• General solution:

�� � = "#� sin��� + 1

$

�%�cos ��� − &�

Vibrations of Continuous Systems

Another Example

• Discrete masses on an elastic string with tension ':

• Consider transverse displacements:

• Vertical force on one mass:

'

'

(�

(�

)� = ' sin (� − ' sin (�= ' (� − (�

= 'ℓ +��� − +� − +� − +��

ℓ

Another Example

• Equation of motion for mass �:

�+�� = )� ='ℓ +��� − +� − +� − +��

+�� + 2 �� �+� − �� � +��� + +�� = 0�� � = '

�ℓ• Normal modes:

+�,� � = ��,� cos ��� − &�

+�

Masses on a String

First normal mode

Second normal mode

Continuous Systems

• What happens when the number of masses goes to

infinity, while the linear mass density remains constant?

�+�� ='ℓ +��� − +� − +� − +��

�ℓ → -

./01./ℓ → 2.

23 3�∆3././51

ℓ → 2.23 3

-ℓ6�+6�� = '

6+6� 3�∆3

− 6+6� 3

Continuous Systems

- 6�+6�� = '

6+6� 3�∆3

− 6+6� 3

ℓ

- 6�+6�� = '

6�+6��

6�+6�� =

-'6�+6��

6�+6�� =

17�6�+6��

The Wave Equation: 7 = '/-

Solutions

• When we had masses, the solutions were

+�,� � = ��,� cos ��� − &�– � labels the mass along the string

– With a continuous system, � is replaced by �.

• Proposed solution to the wave equation for the

continuous string:

+ �, � = 9(�) cos��• Derivatives:

6�+6�� = −�

�9(�) cos��6�+6�� =

6�96�� cos��

Solutions

• Substitute into the wave equation:

6�+6�� =

17�6�+6��

6�96�� = −

��7� 9(�)

6�96�� +

��7� 9 � = 0

• This is the same differential equation as for the

harmonic oscillator.

• Solutions are 9 � = � sin ��/7 + < cos ��/7

Solutions

9 � = � sin ��/7 + < cos ��/7• Boundary conditions at the ends of the string:

9 0 = 9 = = 09 � = � sin ��/7 where �= 7⁄ = ��

• Solutions can be written:

9� � = �� sin���=

• Complete solution describing the motion of the

whole string:

+� �, � = �� sin���= cos���

Properties of the Solutions

+� �, � = �� sin���= cos���

?� =2=�

�� =��7=

9� =�72=

Forced Oscillations

• One end of the string is fixed, the other end is forced with the function @ � = < cos��.

+ 0, � = < cos��+ =, � = 0

• The wave equation still holds so we expect solutions to be of the form

+ �, � = 9(�) cos��

Forced Oscillations

• This time we can’t constrain 9(�) to be zero at both ends.

• Now, let 9 � = � sin �� + A– The constant � is just �/7.

– We need to solve for � and A• Boundary condition at � = =:

sin �=7 + A = 0 ⇒ �=7 + A = C�

AD = C� −�=7

• Condition at � = 0:

< = �D sin AD

Forced Oscillations

• Amplitude of oscillations:

�D =<

sin C� − �=/7• What does this mean?

– The driving force can excite many normal modes of

oscillation

– When � = C�7/=, the amplitude gets very large

Forced Oscillations

C = 1 C = 2

9 = �/2� 9 = �/2�

�/<

�/<

= = 5�7 = 10�/F