Physics 1501: Lecture 26 Today ’ s Agenda

Physics 1501: Lecture 26Today’s Agenda • Homework #9 (due Friday Nov. 4) • Midterm 2: Nov. 16 • Katzenstein Lecture: Nobel Laureate Gerhard t’Hooft • Friday at 4:00 in P-36 … • Topics • Simple Harmonic Motion – masses on springs • Pendulum • Energy of the SHO

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...

SHM So Far • We showed that (which came from F=ma) has the most general solution 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!

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.

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

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

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

LS LR Lecture 26, Act 1Period • What length do we make the simple pendulum so that it has the same period as the rod pendulum? (a)(b) (c)

  where  = 0 cos(t + ) General Physical Pendulum • Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, that we know where the CM is located and what the moment of inertia I about the axis is. • The torque about the rotation (z) axis for small  is (sin  ≈  ) = - Mgd ≈- MgR z-axis R  x CM d Mg

Lecture 26, Act 2Physical Pendulum • A pendulum is made by hanging a thin hoola-hoop of diameter D on a small nail. • What is the angular frequency of oscillation of the hoop for small displacements ? (ICM = mR2 for a hoop) pivot (nail) (a) (b) (c) D

wire   I Torsion Pendulum • Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. • The wire acts like a “rotational spring”. • When the object is rotated, the wire is twisted. This produces a torque that opposes the rotation. • In analogy with a spring, the torque produced is proportional to the displacement: = -k

wire   where I Torsion Pendulum... • Since = -k=Ibecomes Similar to “mass on spring”, except I has taken the place of m (no surprise)

R R R R Lecture 26, Act 3Period • All of the following pendulum bobs have the same mass. Which pendulum rotates the fastest, i.e. has the smallest period? (The wires are identical) A) D) C) B)

F = -kx where wire a k m z-axis x(t) = Acos(t + )   R x  I x CM  = 0 cos(t + ) d Mg Simple Harmonic Oscillator • Spring-mass system • Pendula • General physical pendulum • Simple pendulum • Torsion pendulum

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 Active Figure

U K E U s -A 0 A Energy in SHM • For both the spring and the pendulum, we can derive the SHM solution using energy conservation. • The total energy (K + U) of a system undergoing SMH will always be constant! • This is not surprising since there are only conservative forces present, hence energy is conserved.

U U K E U x x -A 0 A SHM and quadratic potentials • SHM will occur whenever the potential is quadratic. • Generally, this will not be the case: • For example, the potential betweenH atoms in an H2 molecule lookssomething like this:

U(x) = U(x0 ) + U(x0 ) (x- x0 ) + U (x0 ) (x- x0 )2+.... U U x0 x Definex = x - x0 andU(x0 ) = 0 Then U(x) = U (x0 )x2 x  SHM and quadratic potentials... However, if we do a Taylor expansion of this function about the minimum, we find that for smalldisplacements, the potential IS quadratic: U(x) = 0 (since x0 is minimum of potential)

SHM and quadratic potentials... U(x) = U (x0)x2 Letk = U (x0) Then: U(x) = kx2 U U x0 x x  SHM potential !!

What about Friction? • Friction causes the oscillations to get smaller over time • This is known as DAMPING. • As a model, we assume that the force due to friction is proportional to the velocity.

What about Friction? We can guess at a new solution. With,

What about Friction? What does this function look like? (You saw it in lab, it really works)

What about Friction? There is a cuter way to write this function if you remember that exp(ix) = cos x + i sin x .

Damped Simple Harmonic Motion Active Figure • Frequency is now a complex number! What gives? • Real part is the new (reduced) angular frequency • Imaginary part is exponential decay constant underdamped overdamped critically damped

Driven SHM with Resistance • To replace the energy lost to friction, we can drive the motion with a periodic force. (Examples soon). • Adding this to our equation from last time gives, F = F0 cos(wt)

Driven SHM with Resistance • So we have the equation, • As before we use the same general form of solution, • Now we plug this into the above equation, do the derivatives, and we find that the solution works as long as,

Something more surprising happens if you drive the pendulum at exactly the frequency it wants to go, Driven SHM with Resistance • So this is what we need to think about, I.e. the amplitude of the oscillating motion, • Note, that A gets bigger if Fo does, and gets smaller if b or m gets bigger. No surprise there. • Then at least one of the terms in the denominator vanishes and the amplitude gets real big. This is known as resonance.

b small b middling b large Driven SHM with Resistance • Now, consider what b does, w  w = w0

Dramatic example of resonance • In 1940, turbulent winds set up a torsional vibration in the Tacoma Narrow Bridge 

Dramatic example of resonance • when it reached the natural frequency 

Dramatic example of resonance • it collapsed !  Other example: London Millenium Bridge

Lecture 26, Act 4Resonant Motion • Consider the following set of pendula all attached to the same string A D B If I start bob D swinging which of the others will have the largest swing amplitude ? (A) (B) (C) C

Physics 1501: Lecture 26 Today ’ s Agenda