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Physics 1501: Lecture 25 Today ’ s Agenda

Physics 1501: Lecture 25 Today ’ s Agenda. Homework #9 (due Friday Nov. 4) Midterm 2: Nov. 16 Topics Review of static equilibrium Oscillation Simple Harmonic Motion – masses on springs Energy of the SHO. Approach to Statics:. In general, we can use the two equations

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Physics 1501: Lecture 25 Today ’ s Agenda

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  1. Physics 1501: Lecture 25Today’s Agenda • Homework #9 (due Friday Nov. 4) • Midterm 2: Nov. 16 • Topics • Review of static equilibrium • Oscillation • Simple Harmonic Motion – masses on springs • Energy of the SHO

  2. Approach to Statics: • In general, we can use the two equations to solve any statics problems. • When choosing axes about which to calculate torque, we can be clever and make the problem easy....

  3. Lecture 25, Act 1Statics • A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case. • In which cases does the box tip over ? (a) all (b) 2 & 3 (c) 3 only 3 1 2

  4. Consider the bottom right corner of the box to be a pivot point. Lecture 25, Act 1Solution • We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass. • If the box can rotate in such a way that the center of mass islowered, it will !

  5. Lecture 25, ACT 1Solution • We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass. • Consider the bottom right corner of the box to be a pivot point. • If the box can rotate in such a way that the center of mass islowered, it will !

  6. N f rG rf rN tg tN tf= 0 mg Lecture 25, Act 1Addendum • What are the torques ?? (where do the forces act ?) t goes to zero at critical point t switches sign at critical point t always zero

  7. k m k m k m New topic (Ch. 13) 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). • This oscillation is called Simple Harmonic Motion,and is actually very easy to understand...

  8. F = -kx a k m x SHM Dynamics • At any given instant we know thatF= mamust be true. • But in this case F = -kx and ma = • So: -kx = ma = a differential equation for x(t) !

  9. Try the solution x = Acos(t) SHM Dynamics... define this works, so it must be a solution !

  10. ok SHM Solution • We just showed that (which came from F=ma) has the solution x = Acos(t) . • This is not a unique solution, though. x = Asin(t) is also a solution. • The most general solution is a linear combination of these two solutions! x =Bsin(t)+Ccos(t)

  11. = Ccos(t)+ Bsin(t) where C = Acos() and B = Asin() It works! Derivation: We want to use the most general solution: x = Acos(t+) is equivalent to x = Bsin(t)+Ccos(t) x = Acos(t+) = Acos(t) cos - Asin(t) sin So we can use x = Acos(t+)as the most general solution!

  12. SHM Solution... • Drawing of Acos(t ) • A = amplitude of oscillation T = 2/ A     A

  13. SHM Solution... • Drawing of Acos(t + )     

  14. SHM Solution... • Drawing of Acos(t - /2)  A     = Asin(t) !

  15. j k y = 0 F= -ky m What about Vertical Springs? • We already know that for a vertical spring if y is measured from the equilibrium position • The force of the spring is the negative derivative of this function: • So this will be just like the horizontal case:-ky = ma = Which has solution y = Acos(t + ) where

  16. by taking derivatives, since: xMAX = A vMAX = A aMAX = 2A k m x 0 Velocity and Acceleration Position: x(t) = Acos(t + ) Velocity: v(t) = -Asin(t + ) Acceleration: a(t) = -2Acos(t + )

  17. Lecture 25, Act 2Simple Harmonic Motion • 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)

  18. vMAX=A  = Also: k = m2 k m x Example • A mass m = 2kg on a spring oscillates with amplitude A = 10cm. At t=0 its speed is maximum, and is v = +2 m/s. • What is the angular frequency of oscillation  ? • What is the spring constant k ? Sok = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m

  19. x(t) = Acos(t + ) v(t) = -Asin(t + ) a(t) = -2Acos(t + ) x(0) = 0 = Acos()  = /2 or -/2 v(0) > 0 = -Asin()  < 0 So  = -/2    k sin cos m x 0 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):

  20. x(t) = Asin(t) v(t) = Acos(t) a(t) = -2Asin(t) A x(t) t   k m -A x 0 Initial Conditions... So we find  = -/2!! x(t) = Acos(t -/2) v(t) = -Asin(t -/2) a(t) = -2Acos(t -/2)

  21. m Lecture 25, Act 3Initial Conditions • 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(wt) a(t) = -amax cos(wt) k y (b) v(t) = vmax sin(wt) a(t) = amax cos(wt) d t = 0 (c) v(t) = vmax cos(wt) a(t) = -amax cos(wt) 0 (both vmax and amax are positive numbers)

  22. x(t) = Acos(t + ) v(t) = -Asin(t + ) a(t) = -2Acos(t + ) Energy of the Spring-Mass System We know enough to discuss the mechanical energy of the oscillating mass on a spring. Remember, Kinetic energy is always K = 1/2 mv2 K = 1/2 m (-Asin(t + ))2 We also know what the potential energy of a spring is, U = 1/2 k x2 U = 1/2 k (Acos(t + ))2

  23. E = 1/2 kA2 U~cos2 K~sin2    Energy of the Spring-Mass System Add to get E = K + U 1/2 m (A)2sin2(t + )+1/2 k (Acos(t + ))2 Remember that so, E = 1/2 kA2 sin2(t + ) + 1/2 kA2 cos2(t + ) = 1/2 kA2 [ sin2(t + ) + cos2(t + )] = 1/2 kA2

  24. SHM So Far • The most general solution is x = Acos(t + ) where A = amplitude  = frequency  = phase constant • For a mass on a spring • The frequency does not depend on the amplitude !!! • We will see that this is true of all simple harmonic motion ! • The oscillation occurs around the equilibrium point where the force is zero!

  25. z  L m mg The Simple Pendulum • 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.

  26. z  L where m Differential equation for simple harmonic motion ! d  = 0 cos(t + ) mg The Simple Pendulum... • Recall that the torque due to gravity about the rotation (z) axis is = -mgd. d = Lsin   L for small so  = -mg Lq • But =II=mL2

  27. Lecture 25, Act 4Simple Harmonic Motion • You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1. • Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2. • Which of the following is true: (a)T1 = T2 (b)T1 > T2 (c) T1 < T2

  28. The Rod Pendulum • 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

  29. d I • So =Ibecomes where The Rod Pendulum... • The torque about the rotation (z) axis is= -mgd= -mg{L/2}sinq -mg{L/2}q for small q • In this case z L/2  x CM L d mg

  30. LS LR Lecture 25, Act 25Period • What length do we make the simple pendulum so that it has the same period as the rod pendulum? (a)(b) (c)

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