Physics 116 ELECTROMAGNETISM AND OSCILLATORY MOTION Lecture 2 SHM - PowerPoint PPT Presentation

Physics 116 ELECTROMAGNETISM AND OSCILLATORY MOTION Lecture 2 SHM and circular motion Sept 30, 2011 R. J. Wilkes Email: ph116@u.washington.edu

In case you were not here yesterday: First: I’m not Wilkes! Your lecturer today: Victor Polinger Substituting for R. J. Wilkes (He will return on Monday)

Announcements In case you were not here yesterday: (we’ll have a slide like this one most days) - PHYS 116 Course home page: Visit frequently for updated course info! http://faculty.washington.edu/wilkes/116/ - Physics Dept’s general-information page for 100 courses: http://www.phys.washington.edu/1xx/ - Webassign page (for homework and grades): http://webassign.net/washington/login.html 30-Sept-2011 Physics 116 - Au11 3

All HW Assignments and Grades on WebAssign � Homework solutions are submitted online. � Homework solutions are submitted online. � No work is accepted past due date. � No work is accepted past due date. � Click on assignment � Click on assignment name name to begin working to begin working . � Grades � Grades for homework are stored in lecture section. for homework are stored in lecture section. NOTE: Lab grades stored separately, under their own sections own sections. . NOTE: Lab grades stored separately, under their

In case you were not Other Class Resources here yesterday: • Your Fellow Students – You’re encouraged to study and discuss class together! • Physics Study Center (A-wing Mezzanine, just under our lecture room) – Meeting place / work space for students – Get guidance and help from TAs outside 116 office hours – Look over other textbooks on the same subjects Physics Library (6 th floor of C-wing) • – Best view from a comfy chair on campus! • Class discussion board – https://catalyst.uw.edu/gopost/board/wilkes/23253/ • Your TA – Kyle Armour (take a bow, Kyle) – He will post office hours next week – Email questions to him via class email, ph116@uw.edu

In case you were not “Clickers” are required here yesterday: iCue / H-ITT Clickers (TX-3100) • Required to enter answers in quizzes – Be sure to get radio (RF), not infrared (IR) • We will not use them this week • We’ll practice using them next week, and begin using them for pop quizzes thereafter – Bring your clicker to class every day from Oct 3 onward • Why must we torture you with pop quizzes? – Motivation to attend class (physically and mentally)! – Quizzes are designed to be easy I F you are paying attention • Questions will be about something we just discussed!

7 Today Lecture Schedule Physics 116 - Au11 (up to exam 1) 30-Sept-2011

summary from Oscillation terminology yesterday: • Period T = le ngth of one full cycle (time between peaks) • Frequency f = number of cycles per unit time (units = 1 / time) • Amplitude A = maximum displacement (length) • Graph of oscillating object’s position vs time for T = 1 sec, f = 1 Hz, A = 2 m position along the cycle is the phase of the wave: ( t / T ) = (φ / 2π) so φ = 2π ( t / T ) , or φ = 360 ° ( t / T ) 2 T = 1 sec f = 1 / T 1 Unit of frequency = 0 1 cycle/sec 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 = 1 hertz (Hz) A = 2 m -1 -2 Distance, meters time, seconds

Examples / applications 1. Pulse rate is 70 beats / minute • What is f in Hz ? This is just conversion of units: 1 min = 60 sec so f in Hz (cycles/sec) = (cycles/min)*(min/sec) f = { 70 (1/min) }* { 1/60 (min/sec) } = 70/60 (1/sec=Hz) = 1.17 Hz • What is period T in sec ? connection between f and T : f = 1 / T So in this case T = 1 / f = 1 / 1.17 Hz (cycles/sec) = 0.85 sec 2. Mass on a spring moves with amplitude A and period T • How long does it take to move a net distance A ? Definition of A = distance displaced in ¼ of a cycle (1 cycle takes t = T) so it reaches x = A at time t = T/4 A Note: it does NOT reach A/2 in T/8! A/2 Speed is not constant: x = A sin(2 p { t / T }) t=T x=A/2 when sin(2 p { t / T }) = ½ T/4 here t / T = fraction of a full cycle at time t, 2 p { t / T } = fraction of 2 p = phase of sine curve at t T/12 sin -1 ( ½ ) = 0.52 radians = 0.083*2 p So x=A/2 when { t / T }=0.083, or t = 0.083T = T / 12 sin -1 (x) means “angle whose sine is x” (arcsine, or inverse sine) 30-Sept-2011 Physics 116 - Au11 9

Position vs time in simple harmonic motion • Restoring force (spring-mass example): F = - k x • For Simple Harmonic Motion (SHM), k = constant – force is proportional to displacement • Acceleration a = F/m = -(k/m)*x • Now comes magic (i.e. calculus, beyond our scope): as you know, – a = rate of change of velocity v (“derivative” of v) – v = rate of change of position x (“derivative” of x) ⎛ ⎞ ⎛ ⎞ 2 d dx d x k – In the language of calculus, = → − = → = − ⎜ ⎟ ⎜ ⎟ F ma kx m x ⎝ ⎠ ⎝ ⎠ 2 dt dt m dt – This is a “differential equation” for x vs time phase, 0 - 2 π …whose solution (magic!) is 0 1 2 3 4 5 6 ⎛ ⎞ ⎛ ⎞ 1.5 t = ⎜ π ⎟ ⎜ ⎟ x ( t ) A cos ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ 1 T displacement Where A=displacement at t=0 0.5 And m 0 = π T 2 k -0.5 -1 Graph’s lower axis shows time/T ⎛ ⎞ t -1.5 π ⎜ ⎟ Upper axis shows phase 2 0 0.2 0.4 0.6 0.8 1 ⎝ ⎠ T t / T = fraction of period 30-Sept-2011 Physics 116 - Au11 10

SHM’s connection to circular motion • Circular motion is periodic • Coordinate of a point on a circle is given by sin/cos functions • So x or y coordinate of object moving in a circle has exactly the same mathematical description as object in SHM – Plot the x coordinate of a pin on a turntable – and we… – Get the x coordinate of a mass on a spring with same period and A=R – Here’s a web applet that illustrates this: http://www.ngsir.netfirms.com/englishhtm/SpringSHM.htm 30-Sept-2011 Physics 116 - Au11 11

Angular velocity in SHM and circular motion • We can use the mathematical equivalence of circular motion and SHM to find useful formulas about SHM θ = ω ( t ) t – The angular position of a peg on a turntable is just here θ is the total angle swept out, and ω is the angular velocity – ω (omega) is the rate of change of angle with time • (this assumes we accumulate the total number of radians the peg sweeps out, we do not reset to 0 radians whenever it passes its starting point) – We found that the x coordinate of the peg has the same form as in SHM: ⎛ ⎞ ⎛ ⎞ t = ⎜ π ⎟ ⎜ ⎟ x ( t ) A cos ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ T ⎛ ⎞ 1 π = π ⎜ ⎟ – Notice: 2 2 f ⎝ ⎠ T ( ) = ω ω = π x t ( ) A cos t – So if we define we can write 2 f ω = “angular frequency” – • number of radians per second of rotation, for peg on turntable • Nothing actually rotating in SHM, but we use ω = 2 π f anyway 30-Sept-2011 Physics 116 - Au11 12

Velocity in SHM and circular motion • Recall from PHYS 114: – For uniform (constant ω ) circular motion • peg’s speed is v = R ω x component of v points in – x direction for 0 < θ < π radians • • From triangle shown, we can see v x = - (v sin ( θ ) ) so x component of velocity is v x = -R ω sin ( ω t) • θ v v y • So for SHM and circular motion we find: ⎛ ⎞ 1 π = π ⎜ ⎟ 2 2 f v x ⎝ ⎠ R y T θ ω = π 2 f x θ = ω ( t ) t ⎛ ⎞ ⎛ ⎞ t ( ) = π = ω ⎜ ⎜ ⎟ ⎟ x t ( ) A cos 2 A cos t ⎝ ⎠ ⎝ ⎠ T ( ) = − ω ω v t ( ) A sin t 30-Sept-2011 Physics 116 - Au11 13

Acceleration in SHM and circular motion • Again recall from PHYS 114: – For uniform (constant ω ) circular motion • Centripetal acceleration has magnitude a = v 2 / R =(R ω) 2 / R = R ω 2 • Centripetal acceleration points toward center x component of a points in – x direction, for 0 < θ < π/2 radians • • From triangle shown, we see x component a x = - a cos ( θ ) so x component of acceleration is a x (t)= R ω 2 cos ( ω t) • • So for both SHM and circular motion we find: a x ⎛ ⎞ 1 π = π ω = π ⎜ ⎟ θ 2 2 f 2 f ⎝ ⎠ T R a y θ θ = ω ( t ) t x ( ) = ω a x t ( ) A cos t ( ) = − ω ω v t ( ) A sin t ( ) = − ω ω 2 a t ( ) A cos t 30-Sept-2011 Physics 116 - Au11 14

Phase relationships and max values • Maximum values occur when sin/cos = 1, so ( ) = = ω x t ( ) A cos t max x A ( ) = ω = − ω ω v t ( ) A sin t max v A ( ) = − ω ω = ω 2 2 a t ( ) A cos t max a A • x, v and a have phase relationships determined by their trig functions: cos, -sin and -cos, all with the same value of ( ω t ) so at t = 0, x = +max, v = 0, and a = - max. After t=0, x and v are 90 deg out of phase (¼ cycle shift) x and a are 180 deg out of phase (opposite signs – ½ cycle shift) 1.5 1 0.5 x=Acos(wt) x, v, a 0 v= - Awsin(wt) a= - Aw^2cos(wt) -0.5 -1 -1.5 0 0.25 0.5 0.75 1 t / T = fraction of period 30-Sept-2011 Physics 116 - Au11 15