Simple Harmonic Motion

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Cecil Jeffry Wilson
5 years ago
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1 Pendula

2 Simple Harmonic Motion diff. eq. d 2 y dt 2 =!Ky 1. Know frequency (& period) immediately from diff. eq.! = K 2. Initial conditions: they will be of 2 kinds A. at rest initially y(0) = y o v y (0) = 0 COSINE y(t) = y o cos(!t) B. at equilib. position y(0) = 0 v y (0) = v o SINE y(t) = v o! sin(!t)

3 Simple Harmonic Motion Q: How does the period of oscillation depend on amplitude in simple harmonic oscillation? Doubling the amplitude will... d 2 y dt 2 =!Ky A... quadruple the period B... double the period C... not change the period D... half the period critical aspect of SHM: period independent of amplitude

4 pivot Simple Pendulum L m At the end of a light rod of length L is a mass: the bob. The other end of the rod is fixed to a frictionless pivot. Consider the system in the configuration shown. mg r calculate the following: I about pivot τ about pivot = ml 2 = -mgl sin α = τ/i = -(g/l) sin

5 mg pivot L r m the other one d 2! dt 2 = " g L sin(!) Simple Pendulum At the end of a light rod of length L is a mass: the bob. The other end of the rod is fixed to a frictionless pivot. Consider the system in the configuration shown. calculate the following: I about pivot τ about pivot α = ml 2 = τ/i = -(g/l) sin ω = d/dt α = dω/dt = d 2 /dt 2 = -mgl sin

6 Harmonic Motion? diff. eq. d 2! = "K sin(!) dt 2 trial solution!(t) = sin( Kt) d 2! = "K sin( Kt) dt 2 NOT SIMPLE!K sin(") =!K sin sin( Kt)?! [ ]

7 What s happening? -(g/l)sin() g/l π/2 d 2! dt = " g 2 L sin(!) -g/l π/2 π

8 What s happening? -(g/l)sin() displacement 2 g/l π/2 restoring torque π/2 π less than 2 restoring torque -g/l

9 Way Forward d 2! dt 2 = " g L sin(!) d 2! dt 2 " # g L! Small Oscillations π/2 g/l -g/l consider only small angles < 0.2 rad sin(!) "! π/2 π -(g/l)sin() ~ -(g/l)

10 Simple Harmonic Motion diff. eq. d 2! dt 2 = " g L! 1. Know frequency (& period) immediately from diff. eq.! = g L 2. Initial conditions: they will be of 2 kinds A. at rest initially!(0) =! o! (0) = 0 COSINE!(t) =! o cos("t) B. at equilib. position!(t) =!(0) = 0!(0) = v o L SINE v o L" sin("t)

11 2. Different angular frequencies for pendulums! = g L angular frequency of oscillation 0 (t) t=t/2!(t) =! o cos("t) d! dt = "(t) = #! o$ sin($t) angular frequency of oscillation 0 t=t t=2t t Ω m T/4 Ω(t) -Ω m T/2 T 2T t

12 2. Different angular frequencies for pendulums! = g L angular frequency of oscillation!(t) =! o cos("t) d! dt = "(t) = #! o$ sin($t) angular frequency of oscillation small angle assumption!(t)! 1! " o! 1!(t) = " o # sin(#t) $ " o #! #

13 Second Pendulum You wish to construct a seconds pendulum (one whose period is exactly 1 second) using a light wire and a 5.0 kg bob. How long should the wire be? A 0.25 m B 1.00 m C 1.56 m D 9.80 m E 61.6 m

14 Second Pendulum You wish to construct a seconds pendulum (one whose period is exactly 1 second) using a light wire and a 5.0 kg bob. How long should the wire be? A 0.25 m B 1.00 m C 1.56 m D 9.80 m E 61.6 m T = 1 sec ω = 2π rad/sec = L = g (2! ) 2 = 0.25 m g L

15 You build the pendulum with L=0.25 m of wire, but use a heavier bob: m=10.0 kg. This is twice as heavy as you had planned for. The period will now be A 0.5 sec B 0.71 sec C 1.00 sec D 1.41 sec E 2.00 sec

16 You build the pendulum with L=0.25 m of wire, but use a heavier bob: m=10.0 kg. This is twice as heavy as you had planned for. The period will now be A 0.5 sec B 0.71 sec C 1.00 sec D 1.41 sec E 2.00 sec! = g L does not depend on m

17 pivot Physical Pendulum CM L m A uniform bar of length L and mass m is fixed to a frictionless pivot. Consider the system in the configuration shown. mg r calculate the following: I about pivot = (1/3)mL 2 τ about pivot = -(1/2)mgL sin α = τ/i = -(1.5g/L) sin

18 pivot Physical Pendulum CM L m A uniform bar of length L and mass m is fixed to a frictionless pivot. Consider the system in the configuration shown. d 2! dt 2 " # 3g 2L! d 2! dt = " 3g ~ 2 2L sin(!) calculate the following: I about pivot = (1/3)mL 2 τ about pivot = -(1/2)mgL sin α = τ/i = -(1.5g/L) sin ω = d/dt α = dω/dt = d 2 /dt 2

19 d 2! dt 2 Simple Harmonic Motion diff. eq. = " 3g 2L!!(0) =! o! (0) = 0!(t) =! o cos("t) 1. Know frequency (& period) immediately from diff. eq.! = 3g 2L 2. Initial conditions: they will be of 2 kinds A. at rest initially COSINE B. at equilib. position!(t) =!(0) = 0!(0) = v o L SINE v o L" sin("t)

20 pivot 1 m 0.5 kg Example A uniform 1.0-meter rod whose mass is 500 g is released from an angle of o = 0.1 rad. Write down the angle as a function of time (t).! = 3g 2L = 3.83 rad/s!(t) =! o cos("t)!(t) = 0.1cos(3.83t)

21 m pivot h=r/2 CM Physical Pendulum A circular metal disk of radius R and mass m pivots about a screw h=r/2 from the center. calculate the following: I about pivot I mg h τ about pivot α = -mgh sin = τ/i = -(mgh/i) sin d 2! dt 2 = " # $ mgh!! = mgh I I

22 m pivot h=r/2 CM Physical Pendulum A circular metal disk of radius R and mass m pivots about a screw h=r/2 from the center. calculate the following: I about pivot I=0.5 mr 2 + m(r/2) 2 mg h I = 0.75 mr 2! = mgh I = 2 3 g R

23 pivot L=1 m m=0.5 kg m=0.5 kg Example A uniform 1.0-meter rod has a mass m=0.50 kg. A small bob of the same mass is attached to its end. What is the angular frequency of its oscillation? A 1.76 rad/s B 2.35 rad/s C 2.95 rad/s D 3.13 rad/s E 4.43 rad/s

24 pivot m=0.5 kg L=1 m h=0.75 m m=0.5 kg I = (ml 2 ) + (ml 2 )/3 I = (4/3) ml 2 h = (3/4) L M = 2 m total mass! = Mgh I = g L Example A uniform 1.0-meter rod has a mass m=0.50 kg. A small bob of the same mass is attached to its end. What is the angular frequency of its oscillation? A 1.76 rad/s B 2.35 rad/s C 2.95 rad/s D 3.13 rad/s E 4.43 rad/s

Oscillations. PHYS 101 Previous Exam Problems CHAPTER. Simple harmonic motion Mass-spring system Energy in SHM Pendulums

Oscillations. PHYS 101 Previous Exam Problems CHAPTER. Simple harmonic motion Mass-spring system Energy in SHM Pendulums PHYS 101 Previous Exam Problems CHAPTER 15 Oscillations Simple harmonic motion Mass-spring system Energy in SHM Pendulums 1. The displacement of a particle oscillating along the x axis is given as a function