The Wilberforce Pendulum Misay A. Partnof and Steven C. Richards - PowerPoint PPT Presentation

Differential Equations Student Projects http://online.redwoods.edu/deproj/index.htm; 1/33 The Wilberforce Pendulum Misay A. Partnof and Steven C. Richards � College of the Redwoods (Eureka, California) � email: partnof@hotmail.com, sliver3717@cs.com � � � � �

The Wilberforce Pendulum The Wilberforce Pendulum is a pendulum that couples longitudinal 2/33 and torsional oscillations transferring energy between the vertical and the rotational motions. � � � � � � �

One Degree of Freedom A simple oscillator has one degree of freedom. 3/33 Equilibrium Position x k m � � Force kx � � � It requires only one independent coordinate to describe its motion. That coordinate is a displacement from equilibrium in the above system. � �

Newton’s Second Law Newton’s Second Law F = ma applied to springs is 4/33 − kx = m ¨ x. Dividing by m and letting k/m equal to ω 2 0 yields x + ω 2 ¨ 0 = 0 . A well known solution to the spring problem is � x = A cos ( ω 0 t + φ ) , � where ω 0 is the “natural frequency.” � � � � �

Natural Frequency 5/33 • Natural frequency is a constant frequency of a non-driven, undamped harmonic oscillator. • Frequency is dependent on the spring constant k and mass m . • Frequency is not dependent on initial displacement of mass or initial velocity. � � � � � � �

Two Degrees of Freedom Two degree of freedom systems require two independent coordinates 6/33 that describe their motion. k k ′ k m 1 m 2 Two degrees of freedom: � 1. x 1 is the displacement of mass m 1 from its equilibrium position. � � 2. x 2 is the displacement of mass m 2 from its equilibrium position. � � � �

Normal Frequency • Are there any configurations that in which the two masses can os- 7/33 cillate with the same frequency? These are the normal frequencies of the system • If the masses are oscillating with the same frequency and phase an- gle, then they can be said to be oscillating at a normal frequency. � • If the masses are oscillating at equal frequencies that are 180 o out � � of phase, then they also can be said to be oscillating at a normal � frequency. � � �

First Normal Frequency Spring k ′ will remain slack and can be ignored. 8/33 Equilibrium Position x 1 x 2 k ′ k k m 1 m 2 Forces kx kx � Therefore, the mass will oscillate with their natural frequencies � � k ω 1 = m. � � � � �

First Normal Mode The equations of motion describing the first normal mode are 9/33 x 1 = A cos ( ω 1 t + φ 1 ) x 2 = A cos ( ω 1 t + φ 1 ) . In vector form, � � A cos ( ω 1 t + φ 1 ) � x 1 � � x ( t ) = = . x 2 A cos ( ω 1 t + φ 1 ) � � � � � �

Second Normal Frequency If the displacement of the mass are equal and opposite 10/33 Equilibrium Position x 1 x 2 k k ′ k m 1 m 2 Forces ( k + 2 k ′ ) x ( k + 2 k ′ ) x then the force acting on each mass will be − ( k + 2 k ′ ) x , then � x = ( − kx − 2 k ′ ) x m ¨ � x + k + 2 k ′ � ¨ x = 0 � m 2 = k + 2 k ′ � w 2 . m � �

Second Normal Mode The equations of motion describing the second normal mode are 11/33 x 1 = A cos ( ω 2 t + φ 2 ) x 2 = − A cos ( ω 2 t + φ 2 ) . These can be expressed inn vector form by, � A cos ( w 2 t + φ 2 ) � x 1 � � � x ( t ) = = . x 2 − A cos ( ω 2 t + φ 2 ) � � � � � �

Analyzing The Wilberforce Pendulum 12/33 � � � � In the following discussion, z will be measured from equilibrium in the vertical direction, and θ will be measured from equilibrium in rotation. � � �

The Lagrangian The Lagrangian is the kinetic energy minus the potential energy of 13/33 the system, L = K − U = 1 z 2 + 1 θ 2 − 1 2 kz 2 − 1 2 δθ 2 − 1 2 I ˙ 2 m ˙ 2 ǫzθ Where K = 1 z 2 + 1 2 I ˙ θ 2 , 2 m ˙ � � � and � U = 1 2 kz 2 + 1 2 δθ 2 + 1 2 ǫzθ. � � �

Euler-Lagrange Equations The Euler-Lagrange Equations that minimize the Lagrangian of the 14/33 system are d � ∂L � − ∂L ∂z = 0 , dt ∂ ˙ z and � � ∂L � d − ∂L ∂θ = 0 . ∂ ˙ � dt θ � � � � �

Applying the Euler-Lagrange Equations Recall our Lagrangian. 15/33 L = 1 z 2 + 1 θ 2 − 1 2 kz 2 − 1 2 δθ 2 − 1 2 I ˙ 2 m ˙ 2 ǫzθ Applying the Euler-Lagrange equation, � ∂L � 0 = d − ∂L dt ∂ ˙ z ∂z 0 = d � − kz − 1 � dt ( m ˙ z ) − 2 ǫθ � � z + kz + 1 0 = m ¨ 2 ǫθ. � � � � �

Applying the Euler-Lagrange Equations Similarly, 16/33 d � ∂L � − ∂L ∂θ = 0 ∂ ˙ dt θ becomes θ + δθ + 1 I ¨ 2 ǫz = 0 . � � � � � � �

Assumed Solutions for z and θ Thus, the following system describes the motion. 17/33 z + kz + 1 m ¨ 2 ǫθ = 0 θ + δθ + 1 I ¨ 2 ǫz = 0 We will assume solutions z ( t ) = A 1 cos ( ωt + φ ) � θ ( t ) = A 2 cos ( ωt + φ ) , � since they are well known solutions for harmonic oscillators. � � � � �

Differential Equations Taking first derivatives 18/33 z ( t ) = − A 1 ω sin ( ωt + φ ) ˙ ˙ θ ( t ) = − A 2 ω sin ( ωt + φ ) and second derivatives, z ( t ) = − A 1 ω 2 cos ( ωt + φ ) ¨ θ ( t ) = − A 2 ω 2 cos ( ωt + φ ) . ¨ � By substituting these into � z + kz + 1 m ¨ 2 ǫθ = 0 � θ + δθ + 1 � I ¨ 2 ǫz = 0 � we produce the results on the next slide. � �

Solving the Differential Equation These results are 19/33 m ( − A 1 ω 2 cos ( ωt + φ )) + kA 1 cos ( ωt + φ ) + 1 2 ǫA 2 cos ( ωt + φ ) = 0 I ( − A 2 ω 2 cos ( ωt + φ )) + δA 2 cos ( ωt + φ ) + 1 2 ǫA 1 cos ( ωt + φ ) = 0 . Factoring out cos ( ωt + φ ) and dividing the first equation by m and the second by I produces � k � A 1 + ǫ m − ω 2 2 mA 2 = 0 � ǫ � δ � � I − ω 2 2 IA 1 + A 2 + = 0 . � Because these are the natural frequencies of the uncoupled system, we � may set � k/m = ω 2 δ/I = ω 2 θ . and � z �

Finding Non-Trivial Solutions The equations for which we can solve for A 1 and A 2 are 20/33 z − ω 2 ) A 1 + ǫ ( ω 2 2 mA 2 = 0 ǫ 2 IA 1 + ( ω 2 θ − ω 2 ) A 2 + = 0 . In order to find non-trivial solutions, the determinant of the coefficient � matrix must be equal to zero. � � � ω 2 z − ω 2 ǫ � � � 2 m � = 0 . � � ǫ ω 2 θ − ω 2 � � � � 2 I � � �

Normal Frequencies Expanding the determinant and grouping like terms yields 21/33 ǫ 2 � � θ ) ω 2 + ω 4 − ( ω 2 z − ω 2 ω 2 z ω 2 = 0 . θ − 4 mI Compare this to Aω 2 + Bω + C = 0 . Using the quadratic formula yields the normal frequencies � � � � z ) 2 + ǫ 2 1 = 1 ω 2 ω 2 θ + ω 2 ( ω 2 θ − ω 2 z + � 2 mI � � � � z ) 2 + ǫ 2 2 = 1 ω 2 ω 2 θ + ω 2 ( ω 2 θ − ω 2 . � z − 2 mI � � �

Amplitudes We need to know the relation of the amplitudes in both θ and z 22/33 directions when the system is in a normal mode. Since we have chosen ω ’s that make z − ω 2 ) A 1 + ǫ ( ω 2 2 mA 2 = 0 ǫ 2 IA 1 + ( ω 2 θ − ω 2 ) A 2 + = 0 dependent, we only need to solve one for the ratio of the amplitudes. � � � � � � �

Finding the First Amplitude Ratio To solve, z − ω 2 ) A 1 + ǫ 23/33 ( ω 2 2 mA 2 = 0 (1) for the ratio of the amplitudes at the first normal frequency, we substi- tute ω 2 z = ω 2 θ = ω 2 in � � � z ) 2 + ǫ 2 1 = 1 ω 2 ω 2 θ + ω 2 ( ω 2 θ − ω 2 z + 2 mI � to get � ǫ ω 2 1 = ω + . √ � 4 mI We then substitute this result into Equation (1) for ω 2 to get, � � ǫ A 1 + ǫ 2 mA 2 = 0 . − √ � 4 mI �

Finding the Amplitude ratios Solving 24/33 ǫ A 1 + ǫ 2 mA 2 = 0 − √ 4 mI for A 2 /A 1 , we find that the ratio of the amplitudes at ω 1 to be � m r 1 = A 2 = I . A 1 � � � � � � �

Second Amplitude Ratio Using the same procedure, we get 25/33 � m r 2 = A 2 = − I . A 1 Therefore, the amplitude vectors can be written as     A (1) A (1) 1 1 A (1) =  =        A (1) r 1 A (1) 2 1 �  A (2)   A (2)  � 1 1 A (2) =  =  ,     �   A (2) r 2 A (2) � 2 1 � where the superscript denotes the normal frequency at which the am- plitude was obtained. � �

Normal Modes The solution vectors can be written as 26/33  A (1)  z (1) ( t )   1 cos ( ω 1 t + φ 1 ) x (1) =  =  = first mode     θ (1) ( t ) r 1 A (1) 1 cos ( ω 1 t + φ 1 )   A (2) z (2) ( t )   1 cos ( ω 2 t + φ 2 ) x (2) =  = �  = second mode     θ (2) ( t ) r 2 A (2) � 1 cos ( ω 2 t + φ 2 ) � � � � �

General Solutions The general solution can be written as a linear combination of the 27/33 normal modes. z ( t ) = z (1) ( t ) + z (2) ( t ) = A (1) 1 cos ( ω 1 t + φ 1 ) + A (2) 1 cos ( ω 2 t + φ 2 ) θ ( t ) = θ (1) ( t ) + θ (2) ( t ) = A (1) 1 r 1 cos ( ω 1 t + φ 1 ) + A (2) 1 r 2 cos ( ω 2 t + φ 2 ) � � � � � � �