1. 1 A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude). At which position is the magnitude of the acceleration of the mass maximum? A)0 B) M C) E D) a is constant everywhere Clicker Question Room Frequency BA Acceleration = (Net Force/mass) = -(k/m)x x

2. CAPA assignment #13 is due on Friday at 10 pm. CAPA assignment #14 is in the distribution boxes and is due Friday December 2 nd at 10pm. Next week (11/28-12/2) in Section: Lab #6 (with prelab) Finish reading Chapter 11 on Vibrations and Waves 10 am Lecture was videotaped by mistake!!! Video files were deleted. Announcements 2

3. 3 A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude). At what position is the magnitude of the net force on the mass a maximum? A)0 B) M C) E D) force is constant everywhere. Clicker Question Room Frequency BA Net force = -kx x

4. 4 A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude). At what position is the total mechanical energy (PE + KE) a maximum? A) 0 B) M C) E D) energy is constant everywhere. Clicker Question Room Frequency BA No friction or dissipative forces! Energy is conserved. x

5. 5 Energy Analysis of Horizontal Spring and Mass Total Mechanical Energy PE for Spring with equilibrium position at x = 0 is As usual, KE =, so Once oscillator is set in motion, total energy is a constant, hence we can always determine x from v or v from x if we know E T

6. 6 Special Points of Horizontal Spring and Mass Oscillation Amplitude A = E x = A, v = 0x = 0, v = ±v max At turning points x = ±A, v = 0, E T = only PE= At equilibrium point x = 0, v =±v max, E T = only KE= x = -A, v = 0

7. 7 Amplitude - Max speed Relation Amplitude A = E x = A, v = 0x = 0, v = ±v max Total energy is conserved!!! So we can conclude Solving for v max in terms of A gives the result x = -A, v = 0 Related to acceleration kx/m

8. 8 A stiff spring and a floppy spring have potential energy diagrams shown below. Which is the stiff spring? A) PINKB) YELLOW I take two identical masses; one is attached to the stiff spring; the other to the floppy spring. Both are positioned at x = 0 and given the same initial speeds. Which spring produces the largest amplitude motion? A) The stiff spring B) The floppy spring C) Same! Clicker Question Room Frequency BA PE = ½kx 2

9. 9 A stiff spring and a floppy spring have potential energy diagrams shown below. Now the identical masses on the two different springs are pulled to the side and released from rest with the same initial amplitude. Which spring produces the largest maximum speed of its mass? A) The stiff spring B) The floppy spring C) Same! Clicker Question Room Frequency BA PE = ½kx 2

10. 10 Example of Energy Analysis I You are given that a mass m attached to a spring with spring constant k was released from rest from position x 0 (the spring’s equilibrium position is x = 0). What is the speed at position x 1 in between 0 and x 0 ? Are we given the amplitude? Yes! A = x 0 because it was released from rest. Total energy = ½kA 2 and it is a constant! Solve for v and find For final result, plug in x 1

11. 11 Example of Energy Analysis II Note what we can do with a little algebraic manipulation… Units are correct and at x = 0, x = A, we get the correct results! If you are given v at x = 0, you know v max, and you can find a formula for x given some v with magnitude less than v max … If given a v at a particular x you can always find A and v max

12. 12 A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude). What is the speed at the position x = E/2 in terms of v max ? A)v max B)v max /2 C)v max /Sqrt(2) D)v max Sqrt(3/2) E) v max Sqrt(3)/2 Clicker Question Room Frequency BA v = v max sqrt[1-(E 2 /4E 2 )] = v max sqrt[1-(1/4)] x

13. 13 Harmonic Time Dependence of SHM I To find the time dependence of SHM we use an amazing fact: The time dependence of one space component of circular motion is exactly the same as the time dependence of SHM! Consider a particle moving around a circle of radius A in the x-y plane, with constant speed v m A +A –A 0 vmvm  x The angular velocity ω =Δθ/Δt For constant speed, ω is constant and so θ = ωt As usual x = A cos θ so x = A cos ωt at t = 0 x = A sin ωt

14. 14 Harmonic Time Dependence of SHM II How do we relate ω to v m ? A +A –A 0 vmvm  x One cycle takes time T, called the period. SHM has same time dependence, so x = A cos [(v m /A)t] But for SHM, we derived So we have Time to go around circle is 2πA/v m = T = 2π rad/ω Solving yields ω = v m /A

15. 15 SHM Period and Frequencies We just found that T = 2π rad/ω = 2πA/v m. For SHM then Independent of amplitude! Frequency f = 1/T so we find Note even for back and forth motion we have an angular frequency ω! Cosines and sines must be functions of angles; here we will always use radians

16. 16 Simple Harmonic Oscillator with x = 0 at t = 0

17. 17 A mass on a spring oscillates with a certain amplitude and a certain period T. If the mass is doubled, the spring constant of the spring is doubled, and the amplitude of motion is doubled, THEN the period... A) increasesB) decreasesC) stays the same. Clicker Question Room Frequency BA Ratio unchanged if m and k both doubled and the period is independent of amplitude