Lecture 19

1. Lecture 19 Goals: • Chapter 14 • Periodic (oscillatory) motion. • Assignment • No HW this week. • Wednesday: Read through Chapter 15.4

2. Periodic Motion is everywhere Examples of periodic motion • Earth around the sun • Elastic ball bouncing up and down • Quartz crystal in your watch, computer clock, iPod clock, etc.

3. Periodic Motion is everywhere Examples of periodic motion • Heart beat In taking your pulse, you count 70.0 heartbeats in 1 min. What is the period, in seconds, of your heart's oscillations? Period is the time for one oscillation T= 60 sec/ 70.0 = 0.86 s

4. Simple Harmonic Motion (SHM) • We know that if we stretch a spring with a mass on the end and let it go the mass will, if there is no friction, ….do something 1. Pull block to the right until x = A 2. After the block is released from x = A, it will A: remain at rest B: move to the left until it reaches equilibrium and stop there C: move to the left until it reaches x = -A and stop there D: move to the left until it reaches x = -A and then begin to move to the right k m -A 0(≡Xeq) A

5. Simple Harmonic Motion (SHM) • The time it takes the block to complete one cycle is the period T and is measured in seconds. • The frequency, denoted f, is the number of cycles that are completed per unit of time: f = 1 / T. In SI units, f is measured in inverse seconds, or hertz (Hz). • If the period is doubled, the frequency is A. unchanged B. doubled C. halved

6. Simple Harmonic Motion • Suppose that the period is T. • Which of the following points on the t axis are separated by the time interval T? A. K and L B. K and M C. K and P D. L and N E. M and P time

7. Simple Harmonic Motion • Now assume that the t coordinate of point K is 0.25 s. • What is the period T , in seconds? • How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? time

8. Simple Harmonic Motion • Now assume that the t coordinate of point K is 0.25 s. • What is the period T , in seconds? T = 1 s • How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? time

9. Simple Harmonic Motion • Now assume that the t coordinate of point K is 0.25 s. • What is the period T , in seconds? T = 1 s • How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? t = 0.5 s time

10. k m T = 1 s -A 0(≡Xeq) A k 2m T is: T > 1 s T < 1 s T=1 s -A 0(≡Xeq) A

11. k m T = 1 s -A 0(≡Xeq) A 2k m T is: T > 1 s T < 1 s T=1 s -A 0(≡Xeq) A

12. F= -k x a k m x SHM Dynamics: Newton’s Laws still apply • At any given instant we know thatF= mamust be true. • But in this case F = -k x and ma= F • So: -k x = ma =m d2x/dt2 d2x/dt2=-(k/m)x a differential equation for x(t) ! “Simple approach”, guess a solution and see if it works!

13. T = 2/ A    A SHM Solution... • Try cos ( t) • Below is a drawing of A cos ( t) • where A=amplitude of oscillation • [with w = (k/m)½ and w = 2p f = 2p /T ] • T=2 (m/k)½

14. SHM Solution... The general solution is: x(t) = A cos ( wt + f) Use “initial conditions” to determine phase  ! k at t=0 m x(t) = A cos (wt + 0) -A 0(≡Xeq) A k at t=0 m x(t) = A cos (wt + π) -A 0(≡Xeq) A

15. Energy of the Spring-Mass System We know enough to discuss the mechanical energy of the oscillating mass on a spring. x(t) = A cos ( wt + f) If x(t) is displacement from equilibrium, then potential energy is U(t) = ½ k x(t)2 = ½ kA2 cos2 ( wt + f) v(t) = dx/dt  v(t) = A w (-sin ( wt + f)) And so the kinetic energy is just ½ m v(t)2 K(t) = ½ m v(t)2 = ½ m(Aw)2 sin2 ( wt + f) Finally, a(t) = dv/dt = -2A cos(t + )

16. Energy of the Spring-Mass System x(t) = A cos( t +  ) v(t) = -A sin( t +  ) a(t) = -2A cos( t +  ) Potential energy of the spring is U = ½ k x2 = ½ k A2 cos2(t + ) The Kinetic energy is K = ½ mv2 = ½ m(A)2 sin2(t+f) And w2 = k / m or k = m w2 K = ½ k A2 sin2(t+f)

17. E = ½ kA2 U~cos2 K~sin2 Energy of the Spring-Mass System So E = K + U = constant =½ k A2 At maximum displacement K = 0 and U = ½ k A2 and acceleration has it maximum At the equilibrium position K = ½ k A2 = ½ m v2 and U = 0

18. SHM So Far • The most general solution isx = A cos(t + ) where A = amplitude  = (angular) frequency  = phase constant • For SHM without friction, • The frequency does notdepend on the amplitude ! • This is true of all simple harmonic motion! • The oscillation occurs around the equilibrium point where the force is zero! • Energy is a constant, it transfers between potential and kinetic

19. The “Simple” Pendulum • 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. S Fy = may = T – mg cos(q) ≈ m v2/L S Fx = max = -mg sin(q) If q small then x  L q and sin(q) q dx/dt = L dq/dt ax = d2x/dt2 = L d2q/dt2 so ax = -g q = L d2q / dt2 and q = q0 cos(wt + f) with w = (g/L)½ z y  L x T m mg

20. The shaker cart • 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. • At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground. • What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude. Hint: At equilibrium, both the cart and the bag are moving at their maximum speed. By dropping the bag at this point, energy (specifically the kinetic energy of the bag) is lost from the spring-cart system. Thus, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must decrease

21. The shaker cart • 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. • What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude. • Hint: Dropping the bag at maximum distance from equilibrium, both the cart and the bag are at rest. By dropping the bag at this point, no energy is lost from the spring-cart system. Therefore, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must remain constant.

22. The shaker cart • What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart? It increases the maximum speed. It decreases the maximum speed. It has no effect on the maximum speed. Hint: Dropping the bag at maximum distance from equilibrium, both the cart and the bag are at rest. By dropping the bag at this point, no energy is lost from the spring-cart system. Therefore, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must remain constant.