1. Physics 201: Lecture 29, Pg 1 Lecture 29 Goals Goals  Describe oscillatory motion in a simple pendulum  Describe oscillatory motion with torques  Introduce damping in SHM  Discuss resonance 

2. Physics 201: Lecture 29, Pg 2 Final Exam Details l Sunday, May 13th 10:05am-12:05pm in 125 Ag Hall & quiet room l Format:  Closed book  Up to 4 8½x1 sheets, hand written only  Approximately 50% from Chapters 13-15 and 50% 1-12  Bring a calculator l Special needs/ conflicts: All requests for alternative test arrangements should be made by Thursday May10th (except for medical emergency)

3. Physics 201: Lecture 29, Pg 3 Mechanical Energy of the Spring-Mass System Kinetic energy is always K = ½ mv 2 = ½ m(  A) 2 sin 2 (  t+  ) Potential energy of a spring is, U = ½ k x 2 = ½ k A 2 cos 2 (  t +  ) And  2 = k / m or k = m  2 K + U = constant x(t) = A cos(  t +  ) v(t) = -  A sin(  t +  ) a(t) = -  2 A cos(  t +  )

4. Physics 201: Lecture 29, Pg 4 SHM is a close as Lake Mendota… So can you estimate the characteristic frequency for a bobbing in the water? If you have equilibrium and there is a linear restoring force, then yes with  = (k / m) ½ Fmg B y0y0

5. Physics 201: Lecture 29, Pg 5 SHM is a close as Lake Mendota… Deeper than y 0 means and a net force of A linear restoring force with k =  w Ag and boat mass m = Ay 0  w so  = (g / y 0 ) ½ Lighter boats bob more quickly than heavy ones (if the same size) FBFB mg y

6. Physics 201: Lecture 29, Pg 6 Example of phase y = A cos(  t +  ) You have identical vertical springs with identical masses. Both are undergoing simple harmonic motion with frequency f = 1/2  (k/m) ½ l The 1 st mass always moves up when the 2nd mass is moves down. Vertical displacement  0 =t=t T/2 T time 1 st mass 2 nd mass What is the phase difference between the two masses? A: 0 B:  /2 C:  D: 3  /2 E: 2 

7. Physics 201: Lecture 29, Pg 7 The shaker cart l You stand inside a small cart attached to a heavy-duty spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you. l At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground. l What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation? A. It increases the amplitude. B. It decreases the amplitude. C. It has no effect on the amplitude. Hint: At equilibrium, both the cart and the bag are moving at their maximum speed.

8. Physics 201: Lecture 29, Pg 8 The shaker cart l Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium. l What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation? A. It increases the amplitude. B. It decreases the amplitude. C. It has no effect on the amplitude. Hint: At maximum displacement there is no kinetic energy.

9. Physics 201: Lecture 29, Pg 9 The shaker cart l What effect does jettisoning the sandbag at the cart’s maximum displacement from equilibrium have on the maximum speed of the cart? A. It increases the maximum speed. B. It decreases the maximum speed. C. It has no effect on the maximum speed. Hint: At maximum displacement there is no kinetic energy.

10. Physics 201: Lecture 29, Pg 10 The Pendulum (using torque) l A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.   z = I  z = -mg sin(  ) L   z ≈ mL 2  z ≈ -mg  L L (d 2  /dt 2 ) = -g  compare to ma x = -kx d 2  / dt 2 = (-g/L)  with  (t)=  0 cos(  t +  ) and  =(g/L) ½ z  L mg y x T L sin 

11. Physics 201: Lecture 29, Pg 11 The Pendulum l A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements. If  small then sin(  )   0 ° tan 0.00 = sin 0.00 = 0.00 5 ° tan 0.09 = sin 0.09 = 0.09 10 ° tan 0.17 = sin 0.17 = 0.17 15 ° tan 0.26 = 0.27 sin 0.26 = 0.26  L m mg z y x T

12. Physics 201: Lecture 29, Pg 12 The “Simple” Pendulum l A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.  F y = ma y = T – mg cos(  ) = ma c = m v T 2 /L  F x = ma x = -mg sin(  ) where x = L tan  If  small then x  L  and sin(  )   dx/dt = L d  /dt a x = d 2 x/dt 2 = L d 2  /dt 2 so a x = -g  = L d 2  / dt 2  L d 2  / dt 2 - g  = 0 and  =   cos(  t +  ) or  =   sin(  t +  ) with  = (g/L) ½  L m mg z y x T

13. Physics 201: Lecture 29, Pg 13 What about Vertical Springs? l For a vertical spring, if y is measured from the equilibrium position l Recall: force of the spring is the negative derivative of this function: l This will be just like the horizontal case: -ky = ma = j k m F= -ky y = 0 Which has solution y(t) = A cos(  t +  ) where

14. Physics 201: Lecture 29, Pg 14 Exercise Simple Harmonic Motion l A mass oscillates up & down on a spring. It’s position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ? Remember: velocity is slope and acceleration is the curvature t y(t) (a) (b) (c)

15. Physics 201: Lecture 29, Pg 15 Example l A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0 its speed is at a maximum, and is v=+2 m/s  What is the angular frequency of oscillation  ?  What is the spring constant k ? General relationships E = K + U = constant,  = (k/m) ½ So at maximum speed U=0 and ½ mv 2 = E = ½ kA 2 thus k = mv 2 /A 2 = 2 x (2) 2 /(0.1) 2 = 800 N/m,  = 20 rad/sec k x m

16. Physics 201: Lecture 29, Pg 16 Example Initial Conditions l A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction): k m y 0 d (A) v(t) = - v max sin(  t ) a(t) = -a max cos(  t ) (B) v(t) = v max sin(  t ) a(t) = a max cos(  t ) (C) v(t) = v max cos(  t ) a(t) = -a max cos(  t ) (both v max and a max are positive numbers) t = 0

17. Physics 201: Lecture 29, Pg 17 Exercise Initial Conditions l A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction): k m y 0 d (A) v(t) = - v max sin(  t ) a(t) = -a max cos(  t ) (B) v(t) = v max sin(  t ) a(t) = a max cos(  t ) (C) v(t) = v max cos(  t ) a(t) = -a max cos(  t ) (both v max and a max are positive numbers) t = 0

18. Physics 201: Lecture 29, Pg 18 Exercise Simple Harmonic Motion l You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T 1. l Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T 2. Which of the following is true recalling that  = (g/L) ½ (A) T 1 = T 2 (B) T 1 > T 2 (C) T 1 < T 2

19. Physics 201: Lecture 29, Pg 19 A Rod Pendulum l A pendulum is made by suspending a thin rod of length L and mass M at one end. Find the frequency of oscillation for small displacements.  z = I  = -| r x F | = (L/2) mg sin(  ) I rod at end = mL 2 /3 - mL 2 /3   L/2 mg  -1/3 L d 2  /dt 2 = ½ g    L mg z x CM T

20. Physics 201: Lecture 29, Pg 20 General Physical Pendulum Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, that we know where the CM is located and what the moment of inertia I about the axis is. The torque about the rotation (z) axis for small  is (sin  )  = -MgR sin   -MgR    Mg z-axis R x CM where  =  0 cos(  t +  )  

21. Physics 201: Lecture 29, Pg 21 Torsion Pendulum Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. l The wire acts like a “rotational spring”.  When the object is rotated, the wire is twisted. This produces a torque that opposes the rotation.  Torque is proportional to the angular displacement:  = -   where   is the torsion constant   = (  /I) ½ I wire  

22. Physics 201: Lecture 29, Pg 22 Exercise Period All of the following torsional pendulum bobs have the same mass and radius with  = (  /I) ½ l Which pendulum rotates the slowest (i.e. has the longest period) if the wires are identical? RRRR (A) (B) (C) (D)

23. Physics 201: Lecture 29, Pg 23 What about Friction? A velocity dependent drag force (A model) We can guess at a new solution. With, and now  0 2 ≡ k / m Note

24. Physics 201: Lecture 29, Pg 24 What about Friction? A damped exponential if

25. Physics 201: Lecture 29, Pg 25 Variations in the damping Small damping time constant (m/b) Low friction coefficient, b << 2m Moderate damping time constant (m/b) Moderate friction coefficient (b < 2m)

26. Physics 201: Lecture 29, Pg 26 Damped Simple Harmonic Motion l A downward shift in the angular frequency l There are three mathematically distinct regimes underdamped critically damped overdamped

27. Physics 201: Lecture 29, Pg 27 Exercise l Damped oscillations: A can of coke is attached to a spring and is displaced by hand (m = 0.25 kg & k = 25.0 N/m) The coke can is released, and it starts oscillating with an amplitude of A = 0.3 m. How damped is the system? A. Underdamped (multiple oscillations with an exponential decay in amplitude) B. Critically damped (simple decaying motion with at most one overshoot of the system's resting position) C. Overdamped (simple exponentially decaying motion, without any oscillations)

28. Physics 201: Lecture 29, Pg 28 Driven SHM with Resistance Apply a sinusoidal force, F 0 cos (  t), and now consider what A and b do,   b/m small b/m middling b large   Not Zero!!! steady state amplitude

29. Physics 201: Lecture 29, Pg 29 For Thursday l Review for final!