Physics 151 Lecture 31 Todays Agenda l Todays

Physics 151: Lecture 31 Today’s Agenda l Today’s Topics ç Simple Harmonic Motion – masses on springs (Ch 15. 1 -2) ç Energy of the SHO – Ch. 15. 3 ç Pendula. Physics 151: Lecture 31, Pg 1

See text: 15. 1 -2 l 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 oscillate back and forth (if there is no friction). k l This oscillation is called Simple Harmonic Motion, and is actually very easy to understand. . . k k m m m Physics 151: Lecture 31, Pg 2

See text: 15. 1 -2 SHM Dynamics l l At any given instant we know that F = ma must be true. But in this case F = -kx k a m and ma = l So: -kx = ma = x a differential equation for x(t) ! Physics 151: Lecture 31, Pg 3

See text: 15. 1 -2 SHM Dynamics. . . define Try the solution x = Acos( t) this works, so it must be a solution ! Physics 151: Lecture 31, Pg 4

See text: 15. 1 -2 SHM Dynamics. . . But wait a minute. . . what does angular frequency have to do with moving back & forth in a straight line ? ? l y = R cos ( t) y 1 1 2 3 x 4 6 5 0 -1 1 2 3 4 6 5 Physics 151: Lecture 31, Pg 5

See text: 15. 1 -2 SHM Solution l We just showed that (which came from F=ma) has the solution x = Acos( t). l This is not a unique solution, though. x = Asin( t) is also a solution. l The most general solution is a linear combination of these two solutions! x = Bsin( t)+ Ccos( t) l This is equivalent to: x = A cos( t+ ) where is called a phase Physics 151: Lecture 31, Pg 6

See text: 15. 1 -2 SHM Solution. . . l l Drawing of Acos( t ) A = amplitude of oscillation T = 2 / A A Physics 151: Lecture 31, Pg 7

See text: 15. 1 -2 SHM Solution. . . l Drawing of Acos( t + ) Physics 151: Lecture 31, Pg 8

See text: 15. 1 -2 SHM Solution. . . l Drawing of Acos( t - /2) A = Asin( t) ! Physics 151: Lecture 31, Pg 9

See text: 15. 2 Velocity and Acceleration x(t) = Acos( t + ) v(t) = - Asin( t + ) a(t) = - 2 Acos( t + ) Position: Velocity: Acceleration: by taking derivatives, since: x. MAX = A v. MAX = A a. MAX = 2 A k m 0 x Physics 151: Lecture 31, Pg 10

Lecture 31, Act 1 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 ? y(t) (a) (c) t (b) Physics 151: Lecture 31, Pg 11

Lecture 31, Act 1 Solution l The slope of y(t) tells us the sign of the velocity since l y(t) and a(t) have the opposite sign since a(t) = - 2 y(t) a<0 v<0 y(t) a<0 v>0 (a) (c) t (b) a>0 v>0 The answer is (c). Physics 151: Lecture 31, Pg 12

See text: 15. 2 Example l A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t=0 its speed is maximum, and is v = +2 m/s. ç What is the angular frequency of oscillation ? ç What is the spring constant k ? v. MAX = Also: k = m 2 So k = (2 kg) x (20 s -1) 2 = 800 kg/s 2 = 800 N/m k m x Physics 151: Lecture 31, Pg 13

See text: 15. 2 Initial Conditions Use “initial conditions” to determine phase ! Suppose we are told x(0) = 0 , and x is initially increasing (i. e. v(0) = positive): x(t) = Acos( t + ) v(t) = - Asin( t + ) a(t) = - 2 Acos( t + ) So = - /2 k cos sin m 0 x Physics 151: Lecture 31, Pg 14

Lecture 31, Act 2 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): (a) v(t) = vmax sin( t) (b) v(t) = vmax cos( t) (c) v(t) = - vmax sin( t) a(t) = amax cos( t) k a(t) = -amax cos( t) t=0 a(t) = -amax cos( t) y m d 0 (both vmax and amax are positive numbers) Physics 151: Lecture 31, Pg 15

See text: 15. 2 Energy of the Spring-Mass System We know enough to discuss the mechanical energy of the oscillating mass on a spring. Remember, x(t) = Acos( t + ) v(t) = - Asin( t + ) a(t) = - 2 Acos( t + ) Kinetic energy is always K = 1/2 mv 2 K = 1/2 m (- Asin( t + ))2 We also know what the potential energy of a spring is, U = 1/2 k x 2 U = 1/2 k (Acos( t + ))2 Physics 151: Lecture 31, Pg 16

See text: 15. 3 Energy of the Spring-Mass System Add to get E = K + U 1/2 m ( A)2 sin 2( t + ) + 1/2 k (Acos( t + ))2 Remember that so, E = 1/2 k. A 2 sin 2( t + ) + 1/2 k. A 2 cos 2( t + ) = 1/2 k. A 2 ( sin 2( t + ) + cos 2( t + )) = 1/2 k. A 2 E = 1/2 k. A 2 U~cos 2 K~sin 2 Physics 151: Lecture 31, Pg 17

See text: 15. 1 to 15. 3 SHM So Far l The most general solution is x = Acos( t + ) where A = amplitude = frequency = phase constant l For a mass on a spring l ç The frequency does not depend on the amplitude !!! ç We will see that this is true of all simple harmonic motion ! The oscillation occurs around the equilibrium point where the force is zero! Physics 151: Lecture 31, Pg 18

See text: 15. 4 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. z L m mg Physics 151: Lecture 31, Pg 19

Aside: sin and cos for small l A Taylor expansion of sin and cos about = 0 gives: and So for <<1, and Physics 151: Lecture 31, Pg 20

See text: 15 -5 l The Simple Pendulum. . . Recall that the torque due to gravity about the rotation (z) axis is = -mgd. d = Lsin L for small so = -mg L l z But = I I = m. L 2 L where m d Differential equation for simple harmonic motion ! mg = 0 cos( t + ) Physics 151: Lecture 31, Pg 21

Lecture 31, Act 3 Simple Harmonic Motion l l You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T 1. Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T 2. ç Which of the following is true: (a) T 1 = T 2 (b) T 1 > T 2 (c) T 1 < T 2 Physics 151: Lecture 31, Pg 22

Lecture 31, Act 3 Solution Standing up raises the CM of the swing, making it shorter ! L 2 L 1 T 1 Since L 1 > L 2 we see that T 1 > T 2 (b) Physics 151: Lecture 31, Pg 23

Lecture 31, Act 4 Simple Harmonic Motion l Two clocks with basic timekeeping mechanism consist of 1) a mass on a string and 2) a simple pendulum. Both have a period of 1 s on Earth. When taken to the moon which one of the statements below is correct ? a) b) c) d) e) the periods of both is unchanged. one of them has a period shorter than 1 s. the pendulum has a period longer than 1 s. the mass-spring system has a period longer than 1 s. both c) and d) are true. Physics 151: Lecture 31, Pg 24

See text: 15. 4 The 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 x. CM L mg Physics 151: Lecture 31, Pg 25

See text: 15. 4 The Rod Pendulum. . . The torque about the rotation (z) axis is l = -mgd = -mg{L/2}sin -mg{L/2} for small z In this case l l d So = I becomes I L/2 x. CM where L d mg Physics 151: Lecture 31, Pg 26

Lecture 31, Act 4 Period l What length do we make the simple pendulum so that it has the same period as the rod pendulum? LS (a) (b) LR (c) Physics 151: Lecture 31, Pg 27

See text: 15 -5 Lecture 31, Act 4 Solution LS LR b) S = R if Physics 151: Lecture 31, Pg 28

Recap of today’s lecture l Simple Harmonic Motion, ç Example, block on a spring ç Energy of SHM ç Pendula are just like block/spring Physics 151: Lecture 31, Pg 29