Ch. 15 - Oscillations

Dan Finkenstadt, fi[email protected]

November 19, 2015

Oscillations

If you want to find the secrets of the universe, think in terms of energy, frequency and vibration. – Nikola Tesla period T in seconds, time to complete one cycle (2π) 1 frequency f = in s−1 or Hz T angular frequency ω = 2π/T = 2πf Simple Harmonic Oscillator Simple Harmonic Motion True for all simple harmonic motion:

I Ingredients: F = ma and Fs = −kx x = A cos (ωt + ϕo)

I Result: a differential equation v = −ωA sin (ωt + ϕo) d2x k 2 = − x dt m 2 a = −ω A cos (ωt + ϕo)

Examples: Active Learning Exercise

2. Pendulum Problem: Draw! rg For a mass on a spring with k = 100 N/m and ω = m = 1.0 kg, suppose that it is set in motion with p ` an amplitude of 10 cm and zero phase constant (as cosine), draw and label one oscillation cycle of 1. Mass on spring 3. Physical pendulum its r r 1. position 4. kinetic energy k mgd ωs = ω = 2. velocity 5. potential energy m pp I 3. acceleration 6. total mech. energy Solution: 4.8 cm

Solution: Solution: 2.6 m/s 12 Hz

Rosetta Stone Active Learning Exercise

Using ω we can calculate 10 more quantities: ω Problem: A simple harmonic oscillator 1. frequency: f = 2π A simple harmonic oscillator consists of a block 2. Period: T = 1 = 2π f ω of mass 45 g attached to a spring of spring 3. Maximum Speed: v = ωA max constant 240 N m−1, oscillating on a frictionless 2 4. Maximum Accel.: amax = ω A surface. If the block is given an initial velocity of 2 5. Maximum Force: Fmax = m ω A 3.5 m/s starting at the equilibrium position, what 6. Maximum K.E.: K = 1 m ω2A2 2 is the maximum displacement of the block from 7. Maximum P.E.: U = 1 m ω2A2 2 its equilibrium position? 1 2 2 8. Total Energy: E = 2 m ω A

9. phase at time t: φ = (ωt + φo) x 10. phase constant: φ = cos−1 o o A

Active Learning Exercise Active Learning Exercise

Problem: A simple harmonic oscillator Problem: A simple harmonic oscillator A simple harmonic oscillator consists of a block of A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant mass 45 g attached to a spring of spring constant 240 N m−1, oscillating on a frictionless surface. If 240 N m−1, oscillating on a frictionless surface. If the block is displaced 3.5 cm from its equilibrium the block is displaced 3.5 cm from its equilibrium position and released so that its initial velocity is position and released so that its initial velocity is zero, what is its maximum velocity? zero, what is its frequency of oscillation? Solution: 0.29 J

Solution: 5.00 s

Active Learning Exercise Wiley 15.14 – Big Hint

Problem: A simple harmonic oscillator A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 N m−1, oscillating on a frictionless q surface. If the block is displaced 3.5 cm from its Solve for ω = k ; Solve simultaneously for A equilibrium position and released with an initial m and φ : velocity of 2.5 m/s, what is its total energy? o Solve :

0.136 = A cos (5.41 (0.55) + φo) 3.130 = −5.41 A sin (5.41 (0.55) + φo)

Active Learning Exercise Active Learning Exercise Problem: Pendulum Problem: Physical Pendulum A simple pendulum has a natural period of 2.50 s. If the pendulum’s length ` is increased by a mass m and the stick factor of four, the new period will be mass M, 1. Derive the magnitude of the distance d from the pivot to c.o.m. 2. Derive the rotational inertia I of the system. Given the red disk has 3. Go and do Wiley 15.41 after class! Damped Simple Harmonic Motion Why do we always ignore air resistance?

I Oscillatory motion can be damped Fnet = ma I It means that in addition to spring like force: −kx − bv = ma , or

Fs = −kx mx¨ + bx˙ + kx = 0

nd I there is also a resistance to velocity: It is a homogeneous 2 order linear differential equation with sol’n: x(t) = eκt Fd = −bv 2 κt κt κt 0 = mκ e + bκe + ke I We put them together and it diminishes the √ −b ± b2 − 4mk perfect oscillatory behavior. κ = 2m

Yes, there’s more Finally

Rearranging But that’s only κ. Recall our actual answer is eκt s s  b 2 k b k  b 2 b κ = ± − − ±i − t − t 2m m 2m m 2m 2m |{z} | {z } decay x(t) = e this is ωt s 2 k  b  b = Ae−bt/2m cos(ω0t + φ) κ = ±i − − r m 2m 2m k b2 ω0 = − m 4m2 Solution: p I ω = 120/.310 = 19.7 rad/s −1 I 1 cycle takes 2πω = 0.319 s, 29 cycles take 9.25 s

Solution:

−(b/2m)t Af = Aoe −¨1 −0.148¨s (9.25 ¡s) = Aoe = .254Ao

Damped Simple Harmonic Motion Active Learning Exercise Damping: Problem: Exercise q 2 b2 A damped oscillator has mass 310 g, spring I ω → ω − o 4m2 constant 120 N/m and damping coefficient −b t/2m I A → Ao e b = 92 g/s. What is the ratio of the amplitude of −b t/m damped oscillations to the initial amplitude after I E → Eo e 29 cycles? I Subscripted quantities represent the values that would occur without damping, e.g.

I The “natural frequency”: ωo I The initial amplitude: Ao I The initial energy: Eo

I The damping comes from dissipation: F = −b v

Solution Continued Resonance

I Most things vibrate at a natural frequency ωo

I When they are “driven” by an outside oscillatory force they vibrate in complicated ways

I When the “drive” frequency ωd is close to or at the natural frequency ωo bad things can happen: https: //www.youtube.com/watch?v=BE827gwnnk4 Active Learning Exercise Problem: Simple Harmonic Motion In the lab Simple Harmonic Motion, the position x[m] vs. time t[s] for a mass at the end of a spring are measured by a motion sensor on the floor. The spring has a spring constant of 4.94 N/m, and the mass at the end of the spring is 0.5 kg. At t = 0, the mass is pulled up 0.1 m from the equilibrium position, x = 0, and released. Assume that the spring is massless, DRAW the best graph, including numbered tick marks on each axis, to predict the position vs. time for the mass. (As detected by a motion sensor on the floor.)