# Lect21 - Physics 211 Lecture 21 Today's Concepts Simple...

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Unformatted text preview: Physics 211 Lecture 21 Today's Concepts: Simple Harmonic Motion: Mass on a Spring Physics 211 Lecture 21, Slide 1 Physics how to add sinusoidal curves like in number 3 how Not due till friday the hw includes pendulums and we have not discussed it!!! just iron out the tricky stuff like how to know which cos/sin equation is needed. .........i actually think im good for once...... how simple harmonic motion is expressed in the unit circle as demonstrated by the prelecture because it was a little confusing. How to find phi, and what exactly it represents verything from the pre-lecture as usual, please! Could we please discuss some awesome violin-related examples? I would really like to blind my violin teacher with science this week Do Jimi Hendrix's fingers break the laws of relativity? Whether or not you had a pleasant Thanksgiving break, and if Bruce will be our end-of-the-semester present. He would make a lot of people happy during the final. if stelzer and mats got themselves into a bought of fisticuffs, who would walk away? (mind you this is to the death) Physics 211 Lecture 21, Slide 3 Physics Physics 211 Lecture 21, Slide 4 Physics Preflight k m -A 0 A x A mass on a spring moves with simple harmonic motion mass as shown. Where is the acceleration of the mass most positive? A) x = -A B) x = 0 C) x = A Physics 211 Lecture 21, Slide 5 k m -A 0 A x A mass on a spring moves with simple harmonic motion as mass shown. Where is the acceleration of the mass most positive? A) x = -A B) x = 0 C) x = A A) The force to the right is greatest when the displacement to the left is the greatest (Hooke's Law). B) x=0 is the equilibrium point. the mass has more kinetic energy at this point so acceleration must be the biggest C) because -kx=ma, the acceleration will be most positive when the restoring force is negative, or at the end of the oscillation at x=A. Physics 211 Lecture 21, Slide 6 Most general solution: x(t) = Acos(ωt-φ) Physics 211 Lecture 21, Slide 7 Physics demo Physics 211 Lecture 21, Slide 8 Physics SHM Dynamics... SHM What does angular frequency ω have to do with moving back & forth in a straight line ? y = R cos θ = R cos (ωt) y 3 1 4 1 2 1 3 x 6 8 7 2 8 θ 5 1 0 -1 π 2 π 7 2π θ 6 4 5 Physics 211 Lecture 21, Slide 9 SHM Solution... Drawing of Acos(ωt) T = 2π/ω A −2π −π A π 2π θ Physics 211 Lecture 21, Slide 10 SHM Solution... Drawing of Acos(ωt - φ) φ −2π −π π 2π θ Physics 211 Lecture 21, Slide 11 SHM Solution... Drawing of Acos(ωt - π/2) = Asin(ωt) π/2 −2π −π π 2π θ Physics 211 Lecture 21, Slide 12 Preflight 1.00 10 00 90 0 80 0 70 0 60 0 50 0 40 0 30 0 20 0 10 0 0.00 0 0.50 -0.50 -1.00 Suppose the two sinusoidal curves shown above are added together. Which of the plots shown below best represents the result? 0.60 0.40 90 0 90 0 90 0 10 00 80 0 80 0 80 0 70 0 70 0 70 0 60 0 50 0 40 0 30 0 -0.20 20 0 0.00 10 0 A) 0 0.20 -0.40 - 0.60 2.00 1.50 1.00 10 00 60 0 50 0 40 0 30 0 -0.50 20 0 0.00 10 0 0.50 0 B) -1.00 -1.50 -2.00 0.6 10 00 60 0 50 0 40 0 30 0 20 0 0 -0.2 10 0 0.2 0 C) 0.4 -0.4 -0.6 Physics 211 Lecture 21, Slide 13 1.00 10 00 90 0 80 0 70 0 60 0 50 0 40 0 30 0 20 0 10 0 0.00 0 0.50 -0.50 -1.00 0.60 0.40 80 0 90 0 10 00 80 0 90 0 10 00 80 0 90 0 10 00 70 0 70 0 70 0 60 0 50 0 40 0 30 0 -0.20 20 0 0.00 10 0 A) 0 0.20 -0.40 - 0.60 2.00 1.50 1.00 60 0 50 0 40 0 30 0 -0.50 20 0 0.00 10 0 0.50 0 B) -1.00 -1.50 -2.00 0.6 0.4 60 0 50 0 40 0 30 0 20 0 0 -0.2 10 0 0.2 0 C) -0.4 -0.6 A) it will never be purely smooth like b and c B) the sum of two sinusoids with the same period is a sinusoid with that common period. C) Since the functions are added together, the frequency will increase. Sine wave applet Physics 211 Lecture 21, Slide 14 We can also see this using Trig We ⎛ a+b⎞ ⎛ a −b ⎞ sin(a ) + sin(b) = 2sin ⎜ ⎟ cos ⎜ ⎟ 2⎠ 2⎠ ⎝ ⎝ Suppose a = ωt + α b = ωt + β α +β ⎛ sin(ωt ) + sin(ωt + φ ) = 2sin ⎜ ωt + 2 ⎝ = A sin (ωt + φ ) ⎞ ⎛α − β ⎞ ⎟ cos ⎜ ⎟ 2⎠ ⎠ ⎝ Physics 211 Lecture 21, Slide 16 Physics Physics 211 Lecture 21, Slide 17 Physics y = A cos(ωt ) ω = 2πf = 2π(5) v y = −ω A sin(ωt ) a y = −ω 2 A cos(ωt ) vmax = ωA amax = ω2 A In this problem, solve for A which makes amax = g Notice: The expressions for vmax and amax are true and are no matter what the initial conditions were. Physics 211 Lecture 21, Slide 18 Physics Preflight k k Case 1 m m -1 Case 2 0 1 x -2 0 2 x In the two cases shown the mass and the spring are identical but the spring the amplitude of the simple harmonic motion is twice as big in Case 2 as in Case 1. How are the maximum velocities in the two cases related: A) Vmax,2 = Vmax,1 B) Vmax,2 = 2Vmax,1 B) 2V C) Vmax,2 = 4Vmax,1 C) 4V Physics 211 Lecture 21, Slide 19 k k Case 1 m m -1 Case 2 0 1 x -2 0 2 x How are the maximum velocities in the two cases related: A) Vmax,2 = Vmax,1 B) Vmax,2 = 2Vmax,1 B) 2V C) Vmax,2 = 4Vmax,1 C) 4V A) angular velocity only depends on the spring constant and the masses, the amplitude does not effect it so the velocities will be the same B) since the setup would be 1/2kx^2=1/2mv^2 the x is proportional to the v C) Spring potential, and therefore kinetic energy/speed, is related to amplitude squared. vmax = ωA Physics 211 Lecture 21, Slide 20 Act A mass oscillates up & down on a spring. Its position as a mass function of time is shown below. At which of the points shown does the mass have positive velocity and negative velocity acceleration? acceleration? y(t) y(t (A) (C) t (B) Physics 211 Lecture 21, Slide 21 dy The slope of y(t) tells us the sign of the velocity since v y = y(t tells dt y(t) and a(t) have the opposite sign since a(t) = -ω2 y(t) y(t and a(t have a(t y(t a<0 v<0 a<0 v>0 y(t) y(t (A) (C) t (B) a>0 v>0 The answer is (C). Physics 211 Lecture 21, Slide 22 Act A mass hanging from a vertical spring is lifted a distance d mass above equilibrium and released at t = 0. Which of the above following describes its velocity and acceleration as a function of time? (a) v(t) = -vmax sin(ωt) (a) v(t sin( a(t) = -amax cos(ωt) cos( (b) v(t) = vmax sin(ωt) v(t sin( a(t) = amax cos(ωt) cos( (c) v(t) = vmax cos(ωt) cos( v(t a(t) = -amax cos(ωt) cos( k t=0 y m d 0 (both vmax and amax are positive numbers) and are Physics 211 Lecture 21, Slide 23 Since we start with the maximum possible displacement at Since t = 0 we know that: we y = d cos(ωt) dy vy = = −ωd sin(ωt ) = −vmax sin(ωt ) dt dv y ay = = −ω2 d cos(ωt ) = − amax cos(ωt ) dt k t=0 y m d 0 Physics 211 Lecture 21, Slide 24 Physics 211 Lecture 21, Slide 25 Physics m M k d Ftot = 0 M mg = kd Physics 211 Lecture 21, Slide 26 Physics m M V k m d M v0 A y=0 -A Physics 211 Lecture 21, Slide 27 Physics Just after Just before m M V k Oscillating m M m kv d M 0 A y=0 -A Use momentum conservation to find speed of platform + clay just after the collision mV = ( m + M )v0 Use energy conservation to find the maximum compression of the spring, which is the amplitude. 1 1212 2 ( m + M ) v0 + kd = kA 2 2 2 Physics 211 Lecture 21, Slide 28 Physics ...
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