Chapter 13

Periodic Motion or Oscillation

Oscillations – motions that repeat themselves

Examples: 9 swinging chandeliers 9 boats bobbing at anchor 9 guitar strings 9 quartz crystals in wristwatches 9 atoms in solids 9 air molecules that transmit sound 9 …

Oscillations in real word are usually damped; that is, the motion dies out gradually, transferring mechanical energy to thermal energy

1 Periodic motion

Any motion that repeat itself at regular intervals is called periodic motion or harmonic motion.

Key characteristics Period (T) – time required for one cycle of periodic motion. Units: s (seconds/cycle) Frequency (f) – number of oscillations that are completed each second. Units: s-1 (cycle/second) that is 1 hertz = 1 Hz = 1 oscillation per second = 1 s-1

1 f = T

Periodic motion – Simple harmonic motion Simple harmonic motion (SHM) – a special case of a periodic motion when the displacement x of the particle (object) from the origin is given as a sinusoidal function of time

⎛ 2π ⎞ x(t) = xm cos⎜ t + φ ⎟ = xm cos(ω t + φ) ⎝ T ⎠

2π ω = = 2π f T

SI unit for ω is the radian per second

2 Examples

⎛ 2π ⎞ x(t) = xm cos⎜ t +φ ⎟ = xm cos(ω t +φ) ⎝ T ⎠

The velocity and acceleration of SHM

The velocity of SHM dx(t) d v(t) = = []x cos(ω t +φ) dt dt m

v(t) = −ω xm sin(ω t + φ)

The acceleration of SHM dv(t) d a(t) = = []− ω x sin(ω t + φ) dt dt m

2 2 a(t) = −ω xm cos(ω t + φ) = −ω x(t)

3 Examples

x(t) = xm cos(ω t + φ)

v(t) = −ω xm sin(ω t + φ)

2 a(t) = −ω xm cos(ω t + φ)

Checkpoint

In simple harmonic motion, the magnitude of the acceleration is: A) constant B) proportional to the displacement C) inversely proportional to the displacement D) greatest when the velocity is greatest E) never greater than g

2 a(t) = −ω xm cos(ω t + φ)

4 The force law for simple harmonic motion

From Newton’s second law F = ma = −m(ω2 x) For a SHM – a restoring force that is proportional to the displacement but opposite in sign

Example: Hook’s law for a spring

2 F = −kx then k = mω

k m ω = T = 2π m k

example Car springs

When a family of four people with a total mass of 200 kg step into their 1200-kg car, the car springs compress 3.0 cm (a) What is the spring constant for the car springs, assuming that they act as a single spring? (b) How far will the car lower if loaded with 300 kg? (c) What are the period and frequency of the car after hitting a bump? Assume the shock absorbers are poor, so the car really oscillates up and down.

5 example Car springs (cont.)

F (200.0kg) ⋅(9.8 m/s2 ) F = kx then k = = = 6.5⋅104N/m x 0.03 m if the car loaded with 300.0 kg F x = = 4.5⋅10−2 m k m (1200.0 + 300.0)kg The period T = 2π = 2π = 0.92s k 6.5⋅104N/m 1 The frequency f = = 1.09Hz T

example Two springs

Suppose that two springs in figure have different spring

constant k1 and k2. What is the frequency of oscillations of the block? 1 k F = −kx f = 2π m

1 k 1 k f = 1 f = 2 1 2π m 2 2π m

Fnet = −k1x − k2 x = −(k1 + k2 )x 1 k + k f = 1 2 and f = f 2 + f 2 2π m 1 2

6 Energy in simple harmonic motion

The potential energy

1 1 U(t) = kx2 = kx2 cos2 (ω t + φ) 2 2 m

The kinetic energy 1 1 K(t) = mv2 = kx2 sin2 (ω t + φ) 2 2 m

The total mechanical energy

1 E = U + K = kx2 2 m

Examples

1 U(t) = kx2 cos2 (ω t +φ) 2 m

1 K(t) = kx2 sin2 (ω t +φ) 2 m

7 example Problem: energy in SHM

An object of mass m on a a horizontal frictionless surface is attached to a spring with spring constant k. The object is

displaced from equilibrium of x0 horizontally and given an

initial velocity of v0 back toward the equilibrium position. (a) What is the frequency of the motion? (b) What is the initial potential energy? (c) What is the initial kinetic energy? (d) What is the amplitude of the oscillation?

example Problem: energy in SHM

Given : m, k, x0 , v0

Find : f , U0 , K0, xm m 1 T = 2π and f = k T kx2 U = 0 0 2 mv2 K = 0 x 0 2 0 kx2 E = U + K = m 0 0 2

8 The simple pendulum

The restoring force F = mg sin(θ ) For small angles s mg F ≈ mgθ = mg = s L L Comparing to F = −kx mg k = L

m L T = 2π = 2π k g

example

An experiment to measure g

L T = 2π g

L g = (2π )2 T 2 for L = 1.0 m 39.5 g = m / s2 T 2

9 Problem: Pendulum

A 2500 kg demolition ball sings from the end of the crane. The length of the swinging segment of cable is 17 m. (a) Find the period of the swinging, assuming that the system can be treated as a simple pendulum (b) Does the period depends on the ball’s mass?

L 17.0m T = 2π = 2π = 8.3s g 9.8m / s2

Example

Two blocks (m=1.0 kg and M=10 kg) and a spring (k=200 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40. What amplitude of simple harmonic motion of the spring-block’s system puts the smaller block on the verge of slipping over the larger block?

Fmax = mamax = μsmg 2 amax = ω xmax 2 then ω xmax = μs g μ g μ g(m + M ) and x = s = s = 0.59m max ω2 k

10 Periodic Motion

Any motion that repeat itself at regular intervals

Restoration force Simple harmonic motion Equations

F = ma = −cx x(t) = x m cos(ω t +φ) x(t) = xm cos(ω t + φ) v(t) = −ω x sin(ω t +φ) c m 2π m ω = T = 2π ω = = 2π f 2 m c T a(t) = −ω x(t)

Spring Energy Pendilum

m 1 2 1 2 mg L F = −kx T = 2π E = mv x + cx F = − x T = 2π k 2 2 L g

Damped Oscillations Real-world always have some dissipative (frictional) forces. The decrease in amplitude caused by dissipative forces is called by damping. Quite often damping forces are proportional to the velocity of the oscillating object.

F = −kx − bv x x Second-order differential equation d 2 x dx m + b + kx = 0 (we need two initial conditions) dt2 dt −bt / 2m x(t) = xme cos(ω't + φ) k b2 ω'= − m 4m2

11 A pendulum + periodic external force

Model: 3 forces • gravitational force • frictional force is proportional to velocity • periodic external force

d 2θ I = τ +τ +τ dt2 g f ext dθ τ = −mgLsin(θ ), τ = −β ,. τ = F cos(ωt) g f dt ext

d 2θ dθ = −ω2 sin(θ ) −α + f cos(ωt) example 1 dt2 0 dt mgL g β F ω2 = = , α = , f = 0 I L mL2 mL2

d 2θ d 2θ dθ = −ω2 sin(θ ) = −ω2 sin(θ ) −α dt 2 0 dt 2 0 dt

1.2 1.2 1.0 θ(0)=1.0 1.0 θ(0)=1.0, α=0.1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0.0 0.0

(t) (t) θ -0.2 θ -0.2 -0.4 -0.4 -0.6 -0.6 -0.8 -0.8 -1.0 -1.0 -1.2 -1.2 0 102030405060 0 1020304050 time time

12 d 2θ = −ω2 sin(θ ) + f cos(ωt) example 2: beats dt 2 0 When the magnitude of the force is very large – the system is overwhelmed by the driven force (mode locking) and the are no beats

When the magnitude of the force is comparable with the magnitude of the natural restoring force the beats

may occur 2.5

2.0 θ(0)=0.2, α=0.0, f=0.2, ω=1.1

1.5

1.0

0.5

0.0

(t) θ -0.5

-1.0

-1.5

-2.0

-2.5 0 20406080100120 time

example 3: resonance

When the frequency of an external force is close to a 2 natural frequency d θ 2 2 = −ω0 sin(θ ) + f cos(ωt) dt 2.5

2.0 θ(0)=0.2, α=0.0, f=0.1, ω=1.0

1.5

1.0

0.5

0.0

(t) θ -0.5

-1.0

-1.5

-2.0

-2.5 0 20 40 60 80 100 120 time

13 Oscillations: The Tacoma bridge (1940)

14