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Unformatted text preview: Lecture 36
P112 Apr 20, 2007 Announcements
Final Exam Agenda for today
Energy in SHM Taylor Expansions Simple Pendulum Energy in SHM- ideal spring
The total energy (K + U) of a system undergoing SHM will always be constant, since there are only conservative forces present, hence K+U energy is conserved. U E -A 0 K U A s Total mechanical Energy in SHM of an ideal spring
Total mechanical energy is conserved:
E tot 1 2 1 2 = mv + kx 2 2 Inserting v(t)=- A sin( t + ) and x(t)= A cos ( t + ) :
1 2 1 2 1 2 2 1 1 mv + kx = kA sin ( t + ) + kA 2 cos 2 ( t + ) = kA 2 2 2 2 2 2 1 E tot = kA 2 2 Question
Using energy considerations, derive a formula that relates the oscillation's maximum speed, vMAX, to its amplitude A. Taylor Expansion
TE is an approximation of a function as a sum of terms calculated from the values of its derivatives General form:
n= 0 f (n ) (a) (x a) n n!
x3 x5 sin(x) x + 3! 5! x2 x4 cos(x) 1 + 2! 4! x7 + ... 7! .... Example: sine and cosine for small angles
x3 x5 x7 sin(x) x + + ... 3! 5! 7! x2 x4 cos(x) 1 + .... 2! 4! If the angle (x) is very small we can simply write: sin x x cos x 1
This is called "Small Angle Approximation" The Simple Pendulum
A pendulum is made by suspending a mass m at the end of a string of length L. Find the angular frequency of oscillation for small displacements.
z L m mg The Simple Pendulum...
Recall that the torque due to gravity about z the rotation (z) axis is = -mgd. d = Lsin L for small so = -mg L But = I , I = mL2
0 cos( d2 mgL = mL 2 dt
2 L d2 = 2 dt 2 = g L
m d mg Differential equation for simple harmonic motion! t+ ) Simple Harmonic Motion
You are sitting on a swing. You get 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 Solution
We have shown that for a simple pendulum =
L g g L Since T= 2 T =2 If we make a pendulum shorter, it oscillates faster (smaller period) Solution
Standing up raises the CM of the swing, making it shorter! Since L1 > L2 we see that T1 > T2 . L1 L2 T1 T2 Short problems
What is the period of a simple pendulum with length 1 m? How long should a simple pendulum be to have a period of 1.0 s? The Rod Pendulum
A pendulum is made by suspending a thin rod of length L and mass m at one end. Find the angular frequency of oscillation for small displacements.
z xCM L mg The Rod Pendulum...
The torque about the rotation (z) axis is = -mgd = -mg(L/2)sin -mg(L/2) for 1 mL2 small I= d 7 } 6I 8 3 L 1 2 d2 mg = mL 2 In this case 2 3 dt So = I becomes
2 z L/2 xCM L d mg d2 = 2 dt where 3g = 2L Period
What length do we make the simple pendulum so that it has the same period as the rod pendulum? LS LR Summary
Small Oscillations: Larger Oscillations: longer period Physical Pendulum
Small oscillations: ...
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