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Lecture 21: Simple Harmonic Motion

Lecture 21: Simple Harmonic Motion. Today’s Concept: Simple Harmonic Motion: Mass on a Spring. Most general solution: x ( t ) = A cos ( ω t - ϕ ). demo. SHM Dynamics. y = R cos θ = R cos ( ω t ).

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Lecture 21: Simple Harmonic Motion

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  1. Lecture 21: Simple Harmonic Motion Today’s Concept: Simple Harmonic Motion: Mass on a Spring

  2. Most general solution: x(t)=Acos(ω t-ϕ)

  3. demo

  4. SHM Dynamics y=R cosθ=R cos(ω t) What does angularfrequencyωhave to do with moving back and forth in a straight line? y 3 1 1 4 1 2 2 8 3 θ 1 5 θ 0 x 7 6 -1 4 6 8 7 5

  5. SHM Solution Drawing ofA cos(ωt) T=2π /ω A θ -π π 2π -2π A

  6. SHM Solution Drawing ofA cos(ωt-ϕ) ϕ θ π -π 2π -2π

  7. SHM Solution Drawing ofAcos(ω t-π /2)= Asin(ω t) π /2 ω t -π π -2π 2π

  8. In the slide titled, "example", the decision to choose the equation "A*cos(w*t+'phi') seemed like an arbitrary decision. Please explain why that equation was chosen over the other general solution equations. The description of the phase angle is a little confusing too. ϕ θ -π π 2π -2π Drawing of Acos(ωt-ϕ)

  9. In the slide titled, "example", the decision to choose the equation "A*cos(w*t+'phi') seemed like an arbitrary decision. Please explain why that equation was chosen over the other general solution equations. The description of the phase angle is a little confusing too. Is this a sine or a cosine?

  10. Announcements • smartPhysics homework deadlines have been reset to 3:30 PM on April 28 (beginning of final exam). You can get 100% credit if you go back and correct ANY problem on the HW from the beginning of the semester! • Last year’s final exam has been posted • Final exam is worth 200 points and is 2 hours: • Monday April 28, 3:30 – 5:30 pm • No Cheat Sheet allowed on final exam: we’ll provide formulas • This year’s final exam will have approximately: • 10 T/F questions, 3 points each = 30 points • 10 scenarios, each with two 5MC and one 3MC question = 180 – 190 points • Total number of points available on exam will exceed 200: you can miss a few without penalty • Scenarios will be posted by April 21.

  11. Any linear combination of sines and cosines having the same frequency will result in a sinusoidal curve with the same frequency.

  12. CheckPoint: Superposition Suppose the two sinusoidal curves shown above are added together. Which of the plots shown below best represents the result? A) B) C)

  13. Suppose: a= ωt+ α b= ωt+ β Can we talk more about the trigonometrical functions? Like in question #2.

  14. ACT A mass oscillates up & down on a spring. Its position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negativeacceleration? y(t) (A) (C) t (B)

  15. The slope of y(t) tells us the sign of the velocity since y(t)anda(t) have the opposite sign since a(t)=-ω2y(t) a < 0v > 0 a < 0v < 0 The answer is (C). y(t) (A) (C) t a > 0v > 0 (B)

  16. ACT A mass hanging from a vertical spring is lifted a distance dabove equilibrium and released at t = 0. Which of the following describes its velocity and acceleration as a function of time? A) v(t) = -vmaxsin(ωt)a(t) = -amaxcos(ωt) B) v(t) = vmaxsin(ωt)a(t) = amaxcos(ωt) k y d C) v(t) =vmaxcos(ωt)a(t) = -amaxcos(ωt) t = 0 m 0 (both vmax and amax are positive numbers)

  17. ACT Since we start with the maximum possible displacement at t = 0 we know that: y = dcos(ωt) k y d t = 0 m 0

  18. Att= 0,y= 0, moving down Use energy conservation to find A

  19. Or similarly

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