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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 ��� − &�
+�
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���
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