Physics 116 ELECTROMAGNETISM AND OSCILLATORY MOTION Lecture 4 - PowerPoint PPT Presentation

Physics 116 ELECTROMAGNETISM AND OSCILLATORY MOTION Lecture 4 Pendulum motion Damped SHM Oct 4, 2011 R. J. Wilkes Email: ph116@u.washington.edu

Announcements • Homework set 1 due Thursday by 5 pm • Suggestion: Check the New York Times Tuesdays = science page day • Clickers: Registration web page is open https://catalyst.uw.edu/webq/survey/wilkes/142773 All registered 116 students got an email yesterday with this link (to your @u address) (if you are not registered for 116, you cannot access the page) if you did not get an email about this, and are registered for the course, see me “Survey questions” capture the info we need to connect your clicker ID to your name Bring your clicker every day from now on • Typo in formula yesterday:

Yesterday: Period for SHM • Now that we know a (t), we can find period T without calculus: { } ( ) − kx = − m A ω 2 cos ω t F = ma → − kx = ma → ⎛ ⎞ k ( ) = − ( ) − A ω 2 cos ω t ⎟ A cos ω t ⎜ So ⎝ ⎠ m ⎛ ⎞ ω 2 = k k → ω = ⎜ ⎟ ⎝ ⎠ m m ⎛ ⎞ Also: this meant “newton-meters”, 1 ω = 2 π f = 2 π • We defined ⎜ ⎟ not a minus sign! ⎝ ⎠ T ⎛ ⎞ 1 m T = 2 π → T = 2 π ⎜ ⎟ • So ⎝ ⎠ ω k • This tells us that T increases for larger m, or smaller k 3-Oct-2011 Physics 116 - Au11 3

Lecture Schedule (up to exam 1) Today 4-Oct-11 Physics 116 - Au11 4

SHM: The story so far… = → − = F ma kx ma ⎛ ⎞ 1 ω = π = π ⎜ ⎟ 2 2 f ⎝ ⎠ T k m ω = → = π T 2 m k = ω ( ) cos( ) x t A t = − ω ω v ( t ) A sin( t ) = − ω ω 2 a ( t ) A cos( t ) 1 1 1 = ω = ω = + = 2 2 2 2 2 U kA cos ( t ), K kA sin ( t ), E U K kA 2 2 2 4-Oct-11 Physics 116 - Au11 5

Pendulum motion as example of SHM • A pendulum oscillates about its rest position IF its max displacement is small, this motion is SHM • SHM occurs only when restoring force is F = -kx • For small angles (you decide what’s small!), θ ≈ θ → θ ≈ θ sin mg sin mg So F is proportional to θ (for larger θ , nonlinear so not exactly SHM) Initial height (max angle) of bob determines E U=mgh (h=initial height relative to rest position) E swaps between K and U as with spring-mass system: K=mgh when bob is at h=0 “Galileo’s pendulum” demo illustrates this… http://paws.kettering.edu/~drussell/Demos/Pendulum/Pendula.html If m of string ~ 0 this is a simple pendulum If we have to worry about m of the string or other support, it’s a physical pendulum 4-Oct-2011 Physics 116 - Au11 6

Examples / applications • Similar triangles for string and vertical, and for mg components So for small displacements, θ = s / L • ⎛ ⎞ mg = − θ = − ⎜ ⎟ Minus sign means: F restoring mg s ⎝ ⎠ Opposite displacement s L ⎛ ⎞ mg = ⎜ ⎟ k ⎝ ⎠ L m L = π → = π T 2 T 2 k g How long must a simple pendulum be to have period T = 2 s? (so it ticks every 1 second) θ s L = π T 2 g ( ) ( ) 2 2 2 ( ) L gT 9 . 8 m / s 2 s = π → = = = 2 2 T 2 L 0 . 994 m ( ) ( ) π π 2 2 g 2 2 4-Oct-11 Physics 116 - Au11 7 7

Physical pendulum (review 114!) • If mass of the bob’s support (string, rod, etc) is not negligible, we have to take into account its moment of inertia – The support itself is an extended object rotating about its pivot • So if it has moment of inertia I about its pivot point l I = π T 2 2 l g m where l = distance of CM of support from P For a simple pendulum, bob = a point mass ( I = m l 2 ), and l = L, so 2 L mL L = π = π 2 2 T 2 g mL g If the pendulum is a solid rod, ( I = m L 2 / 3), θ and CM is located at its center ( l = L / 2) 2 L mL So rod has smaller T than simple l I 2 L 3 2 = π = π = π pendulum of same L T 2 2 2 2 2 l ⎛ ⎞ g m g 3 g L Notice: m doesn’t matter! ⎜ ⎟ m ⎝ ⎠ 2 4-Oct-11 Physics 116 - Au11 8 8

Examples / applications • Physical pendulum: Hoop on a peg – recall: parallel axis theorem R I = I (about some parallel axis) + mR 2 Where R= distance between axes ( ) ( ) = + = 2 2 2 I mR mR 2 mR about center = l R 2 l I R 2 mR 2 R = π = π = π T 2 2 2 2 2 l g m g mR g D = π Notice : T 2 g So period is same as simple pendulum with L=diameter of hoop (but true only for this particular example!) R • Solid disk hung on a peg ( ) ⎛ ⎞ 1 3 = + = ⎜ ⎟ 2 2 2 Now : I mR mR mR ⎝ ⎠ 2 2 about center 2 l I R 3 mR / 2 3 R = π = π = π T 2 2 2 l 2 2 g m g mR 2 g 4-Oct-11 Physics 116 - Au11 9 9

Examples / applications • Physical pendulum: ballistic pendulum – Method for measuring projectile speed – Shoot bullet into wood block suspended by strings 1.45 kg wood block, L=0.745 m, 0.0095 kg bullet v 0 Block rises 0.124 m Strategy: use energy considerations to find initial speed of block Use momentum conservation (review114: inelastic collisions) to find v 0 = + = → = → = E U K const K K K U bullet block block block Always! At time of collision ( initial ) (at max height) 1 = + = = + 2 U ( m M ) gh K ( m M ) v 2 Here, v = speed of bullet+block = = = 2 v 2 gh 2 ( 9 . 8 m / s 0 . 124 m 1 . 56 m / s just after collision = + mv ( m M ) v = Conservation of momentum in the collision 0 + + ( ) ( 0.0095 kg 1.45 kg ) 1.56 m/s m M v = = = (about 860 km/hr) v 240 m/s 0 m 0.0095 kg 4-Oct-11 Physics 116 - Au11 10 10

Examples / applications • Spring-mass vs pendulum – Mass m stretches spring (constant k) by distance L when attached k – What is period of mass-spring system? m L = = → = → = m F kx mg kL mg L spring k g m L = π = π T 2 2 k g L So spring-mass system has m same period as a simple pendulum with L = displacement of mass when attached to spring 4-Oct-11 Physics 116 - Au11 11 11

Damped oscillations • If friction is significant, total energy is not constant E = U + K = ½ kA 2 diminishes with time: oscillation gets damped out Common case: friction force is proportional to speed = − (example: air resistance) F damping b v For this case, amplitude A decreases exponentially with time The exponential function imposes an = − A ( t ) A ( t ) exp( bt / 2 m ) envelope on the oscillations 0 – If damping is too great, we never get even the first half cycle – If damping is just right (critical damping) we get just the 1 st half cycle – Critical damping is a desirable feature in many applications • For example, car springs: shock absorbers (= dampers) are set to provide critical damping for a comfortable ride 4-Oct-11 Physics 116 - Au11 12 12

1.5 Damped oscillations Exponential “envelope” 1 • Underdamped (oscillates displacement 0.5 several cycles) exp(-bt/2m) 0 oscillator -0.5 • Overdamped (quits -1 immediately) Oscillator motion 1.5 -1.5 0 0.5 1 1.5 1 t / T = fraction of period displacement 0.5 exp(-bt/2m) 0 oscillator -0.5 1.5 -1 1 -1.5 displacement 0.5 0 0.1 0.2 0.3 0.4 0.5 exp(-bt/2m) t / T = fraction of period 0 oscillator -0.5 -1 • Critically damped -1.5 0 0.1 0.2 0.3 0.4 0.5 (smooth return to zero) t / T = fraction of period 4-Oct-11 Physics 116 - Au11 13 13