Ch 14 – Oscillatory Motion

Motion of an Object Attached to a Spring

The restoring force (Hooke’s law) is the force applied by a spring to bring the system back to equilibrium.

v = dx/dt

FNet = ma; Fs = -kx

a = dv/dt

ma = -kx

a = d^2x/ dt^2

a = -(k/m) x

d^2x/ dt^2 = -(k/m) x

Harmonic Motion: Simple: OW-A-HS (use apparatus of Driven, below)

Mathematical Representation of Simple Harmonic Motion

d^2x/ dt^2 = -(k/m) x Now we have to find a math function that fits this case. First we’re going to allow ω^2 = k/m why ω^2; we know the result… and it’s best for simplification of our equations if ω^2 = k/m

d^2x/ dt^2 = -(k/m) x d^2x/ dt^2 = - ω^2x

AF 15.2 AF 15.7 AF 15.9

We’ll use the following solution x(t) = A cos(ωt + φ) (sine also works…but we’re choosing cosine)

v = dx/dt = -A ω sin(ωt + φ) a = dv/dt = -A ω^2 cos(ωt + φ) Where φ is the phase constant; ω = √(k/m) is angular velocity (in radians) which is also called angular frequency T = 1/f ω = 2πf

A simple harmonic oscillator takes 12.0 s to undergo five complete vibrations. Find (a) the period of its motion (b) the frequency in hertz (c) angular frequency in radians/sec

(a) T = 12 sec / 5 vib T = 2.4 sec

(b) f = 1/T f = 1 / 2.4 f = 0.417 Hz

(c) ω = 2πf ω = 2π(0.417) ω = 2.62 rad/s

Energy of the Simple Harmonic Oscillator

K = ½ mv^2

U = ½ kx^2

K = ½ m(-A ω sin(ωt + φ))^2 K = ½ m A^2ω^2sin^2(ωt + φ)

U = ½ k (A cos(ωt + φ))^2 U = ½ k A^2cos^2(ωt + φ)

TE = K + U; sin^2θ + cos^2θ = 1; ω^2 = k/m

TE = ½ m A^2ω^2 sin^2(ωt + φ) + ½ k A^2cos^2(ωt + φ) TE = ½ m A^2(k/m) sin^2(ωt + φ) + ½ k A^2cos^2(ωt + φ) TE = ½kA^2

Example A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is 2.00 J, what is the amplitude of the motion?

TE = ½kA^2 A^2 = 2 TE / k A^2 = 2 (2) / 126.3 A = 0.178 meters

ω^2 = k/m; ω = 2πf; f = 1/T 4π^2/T^2 = k/m k = 4 π^2 m / T^2 k = 4 π^2 m / T^2 k = 126.3 N/m

Comparing Simple Harmonic Motion with Uniform Circular Motion

x(t) = A cos(ωt + φ)

The Pendulum

θ < 10°; tan θ = sin θ = θ (θ must be in radians) s = L θ

Newton’s 2^nd law in rotation Στ = I α x = d = sinθ R (lever arm) τ = F•d Στ = I α mg(sinθ R) = I (d^2θ/dt^2) at small angles mgR θ = I (d^2θ/dt^2) d^2θ/dt^2 = (mgR/I) θ in the same form as d^2x/dt^2 = -(k/m) x with the solution of x(t) = A cos(ωt + φ) where ω = √(k/m) θ(t) = θmax cos(ωt + φ) where ω = √(mgR/I); used to measure the moment of inertia T = 2π √(L/g)

Torsional Pendulum Any suspended rod with a mass attached at the bottom, when torqued rotates as a Torsional pendulum where τ = - κθ Στ = I α Στ = I (d^2θ/dt^2) - κθ = I (d^2θ/dt^2) d^2θ/dt^2 = -(κ/I) θ in the same form as d^2x/dt^2 = -(k/m) x θ(t) = θmax cos(ωt + φ) where ω = √( κ /I) T = 2π √(I/ κ)

F = ma

F = -mg sinθ

ma = -mg sinθ a = -g sinθ d^2s/dt^2 = -g sinθ d^2s/dt^2 = -g θ Ld^2θ/dt^2 = -g θ d^2θ/dt^2 = -(g/L) θ in the same form as d^2x/dt^2 = -(k/m) x with the solution of x(t) = A cos(ωt + φ) where ω = √(k/m) θ(t) = θmax cos(ωt + φ) where ω = √(g/L) T = 2π/ω = 2π/√(g/L)

Example A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 12.0 s. How tall is the tower?

T = 2π √(L/g) L = T^2 g /4π^2 L = 12^2 (9.8)/ 4π^2 L = 35.7 meters

Simple Pendulum: OW-A-SP

Damped Oscillations

Damped Oscillator: Amplitude decreases in time until stops.

Natural Frequency: The freq w/o a resistive force, bv

ΣFx = -kx –bvx = m ax section 6.4 -kx –b vx = m ax -kx –b dx/dt = m d^2x/dt^2 solution (when resistive force is insignificant) where or

where ωo^2 = k/m and ωo is called the natural frequency

underdamped

critically damped

overdamped

bvmax < kA

bcrit/2m = ωo

bvmax > kA and b/2m > ωo

Example A pendulum with a length of 1.00 m is released from an initial angle of 15.0°. After 1000 s, its amplitude has been reduced by friction to 5.50°. What is the value of b/2m?

x = A e^-bt/2m

5.5/15 = e^-b(1000)/2m ln(5.5/15) = -1000b / 2m -1.003 = -1000b / 2m b/2m = 0.001 s^-1

xf = Ae^-b(1000)/2m

xi = A e^-bt/2m

5.5 = e^-b(1000)/2m

15 = e^(0)b/2m

Forced Oscillations

Harmonic Motion, Driven OW-A-HM

By applying a constant force that does positive work, we can compensate the loss of amplitude for a damped oscillator. The classic example is the child on a swing…to maintain a large amplitude the child must maintain an energy input by pumping his/her legs. (legs pumping) (resistive force) (Hooke’s law) ΣF = Fcyclic added force – b v - kx ΣF = Fo sin ωt – b dx/dt - kx

Solution is x(t) = A cos(ωt + φ) When ω is approx equal to ωo and when b is small {b (as in –bv) is called the damping term} resonance occurs. Resonance occurs when a small energy input (positive work) leads to a dramatic increase in amplitude near the natural frequency, ωo.

** The below section is copied verbatim from Serway/Jewitt ** The applied force is in phase with the velocity at resonance frequency. The rate at which work is done on the oscillator by F equals the dot product F•v; this rate is the power delivered to the oscillator. Because the product F•v is a maximum when F and v are in phase, we conclude that at resonance the applied force is in phase with the velocity and the power transferred to the oscillator is a maximum.

Example A weight of 40.0 N is suspended from a spring that has a force constant of 200 N/m. The system is undamped and is subjected to a harmonic driving force of frequency 10.0 Hz, resulting in a forced-motion amplitude of 2.00 cm. Determine the maximum value of the driving force.

Since b = 0,

ω = 2πf ω = 2π10 ω = 20 π ω^2 = 4000 s^-2

ωo^2 = k / m ωo^2 = 200/4 kg ωo^2 = 50 rad^2/s^2

A = F /m / (ω^2 - ωo^2) 0.02 = (F/4) /(4000-50) F = 316 N