Harmonic Motion and Waves

Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude – maximum displacement

Period and Frequency Period (T) – Time for one complete cycle (back to starting point) Frequency (Hz) – Cycles per second F = 1 T = 1 T f

Period and Frequency A radio station has a frequency of 103.1 M Hz. What is the period of the wave? 103.1 M Hz 1X106 Hz = 1.031 X 108 Hz 1M Hz T = 1/f = 1/(1.031 X 108 Hz) = 9.700 X 10-9 s

Hooke’s Law F = -kx F = weight of an object k = spring constant (N/m) x = displacement when the object is placed on the spring

Hooke’s Law: Example 1 What is the spring constant if a 0.100 kg mass causes the spring to stretch 6.0 cm? (ANS: 16 N/m)

Special Note: If a spring has a mass on it, and then is stretched further, equilibrium position is the starting length (with the mass on it)

Hooke’s Law: Example 2 A family of four has a combined mass of 200 kg. When they step in their 1200 kg car, the shocks compress 3.0 cm. What is the spring constant of the shocks? F = -kx k = -F/x k = -(200 kg)(9.8 m/s2)/(-0.03 m) k = 6.5 X 104 N/m (note that we did not include the mass of the car)

Hooke’s Law: Example 2a How far will the car lower if a 300 kg family borrows the car? F = -kx x = -F/k x = -(300 kg)(9.8 m/s2)/ 6.5 X 104 N/m x = 4.5 X 10-2 m = 4.5 cm

Forces on a Spring Extreme Position (Amplitude) Equilibrium position Force at maximum Velocity = 0 Equilibrium position Force = 0 Velocity at maximum

Energy and Springs KE = ½ mv2 PE = ½ kx2 Maximum PE = ½ kA2 Law of conservation of Energy ½ kA2 = ½ mv2+ ½ kx2

All PE All KE Some KE and Some PE

Spring Energy: Example 1 A 0.50 kg mass is connected to a light spring with a spring constant of 20 N/m. Calculate the total energy if the amplitude is 3.0 cm. Maximum PE = ½ kA2 Maximum PE = ½ (20 N/m)(0.030 m2) Maximum PE = 9 X 10-3 Nm (J)

Spring Energy: Example 1a What is the maximum speed of the mass? ½ kA2 = ½ mv2+ ½ kx2 ½ kA2 = ½ mv2 (x=0 at the origin) 9 X 10-3 J = ½ (0.50 kg)v2 v = 0.19 m/s

Spring Energy: Example 1b What is the potential energy and kinetic energy at x = 2.0 cm? PE = ½ kx2 PE = ½ (20 N/m)(0.020 m2) = 4 X 10-3 J ½ kA2 = ½ mv2 + ½ kx2 ½ mv2 = ½ kA2 - ½ kx2 KE = 9 X 10-3 J - 4 X 10-3 J = 5 X 10-3 J

Spring Energy: Example 1c At what position is the speed 0.10 m/s? (Ans: + 2.6 cm)

Spring Energy: Example 2a A spring stretches 0.150 m when a 0.300 kg mass is suspended from it (diagrams a and b). Find the spring constant. (Ans: 19.6 N/m)

Spring Energy: Example 2b The spring is now stretched an additional 0.100 m and allowed to oscillate (diagram c). What is the maximum velocity?

The maximum velocity occurs through the origin: ½ kA2 = ½ mv2+ ½ kx2 ½ kA2 = ½ mv2 (x=0 at the origin) kA2 = mv2 v2 = kA2/m v = \/kA2/m = \/(19.6 N/m)(0.100m)2/0.300kg v = 0.808 m/s

Spring Energy: Example 2c What is the velocity at x = 0.0500 m? ½ kA2 = ½ mv2+ ½ kx2 kA2 = mv2+ kx2 mv2 = kA2 - kx2 v2 = kA2 - kx2 m v2 = 19.6 N/m(0.100m2 – 0.0500m2) = 0.49 m2/s2 0.300 kg v = 0.700 m/s

Spring Energy: Example 2d What is the maximum acceleration? The force is a maximum at the amplitude F = ma and F = kx ma = kx a = kx/m = (19.6 N/m)(0.100 m)/(0.300 kg) a = 6.53 m/s2

Trigonometry and SHM Ball rotates on a table Looks like a spring from the side One rev(diameter) = 2pA T = 2p m k f = 1 T

Period depends only on mass and spring constant Amplitude does not affect period vo = 2pAf or vo = 2pA T vo is the initial (and maximum) velocity

Period: Example 1 What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 104 N/m after it hits a bump? T = 2p m = 2p (1400 kg/6.5 X 104 N/m)1/2 k T = 0.92 s f = 1/T = 1/0.92 s = 1.09 Hz

Period: Example 2a An insect (m=0.30 g) is caught in a spiderweb that vibrates at 15 Hz. What is the spring constant of the web? T = 1/f = 1/15 Hz = 0.0667 s T = 2p m k T2 = (2p)2m k = (2p)2m = (2p)2(3.0 X 10-4 kg) = 2.7 N/m T2 (0.0667)2

Period: Example 2b What would be the frequency for a lighter insect, 0.10 g? Would it be higher or lower? T = 2p m k T = 2p (m/k)1/2 T = 2p (1.0 X 10-4 kg/2.7 N/m)1/2 = 0.038 s f = 1/T = 1/0.038 s = 26 Hz

Cosines and Sines Imagine placing a pen on a vibrating mass Draws a cosine wave

x = A cos2pt or x = A cos2pft T A = Amplitude t = time T = period f = frequency

x = A cos2pft v = -vosin2pft a = -aocos2pft Velocity is the derivative of position Acceleration is the derivative of velocity

Cos: Example 1a A loudspeaker vibrates at 262 Hz (middle C). The amplitude of the cone of the speaker is 1.5 X 10-4 m. What is the equation to describe the position of the cone over time? x = A cos2pft x = (1.5 X 10-4 m) cos2p(262 s-1)t x = (1.5 X 10-4 m) cos(1650 s-1)t

Cos: Example 1b What is the position at t = 1.00 ms (1 X 10-3 s) x = A cos2pft x = (1.5 X 10-4 m) cos2p(262 s-1) (1 X 10-3 s) x = (1.5 X 10-4 m) cos(1.65 rad) = -1.2 X 10-5 m

Cos: Example 1c What is the maximum velocity and acceleration? vo = 2pAf vo = 2p(1.5 X 10-4 m)(262 s-1) = 0.25 m/s

F = ma kx = ma a = kx/m But we don’t know k or m a = k x Solve for k/m m T = 2p m k T2 = (2p)2m k = (2p)2 = (2p)2f2 m T2

a = k x m a = (2pf)2x = (2pf)2A a = [(2p)(262 Hz)]2(1.5 X 10-4 m) = 410 m/s2

Cos: Example 2a Find the amplitude, frequency and period of motion for an object vibrating at the end of a spring that follows the equation: x = (0.25 m)cos p t 8.0

x = A cos2pft x = (0.25 m)cos p t 8.0 Therefore A = 0.25 m 2pft = p t 2f = 1 f = 1/16 Hz T = 1/f = 16 s

Cos: Example 2b Find the position of the object after 2.0 seconds. x = (0.25 m)cos p t 8.0 x = (0.25 m)cos p 4.0 x = 0.18 m

The Pendulum Pendulums follow SHM only for small angles (<15o) The restoring force is at a maximum at the top of the swing. q Fr = restoring Force

Remember the circle (360o = 2p rad) q = x L Fr = mgsinq at small angles sinq = q Fr = mgq q L x q mg Fr

Fr = mgq Fr = mgx (Look’s like Hook’s Law F = -kx) L k = mg T = 2p m k T = 2p mL mg

T = 2p L g f = 1 = 1 g T 2p L The Period and Frequency of a pendulum depends only on its length

Swings and the Pendulum To go fast, you need a high frequency Short length (tucking and extending your legs) f = 1 g 2p L decrease the denominator

Example 1: Pendulum What would be the period of a grandfather clock with a 1.0 m long pendulum? T = 2p L g Ans: 2.0 s

Example 2: Pendulum Estimate the length of the pendulum of a grandfather clock that ticks once per second (T = 1.0 s). T = 2p L g Ans: 0.25 m

Damped Harmonic Motion Most SHM systems slowly stop For car shocks, a fluid “dampens” the motion

Resonance: Forced Vibrations Can manually move a spring (sitting on a car and bouncing it) Natural or Resonant frequency (fo) When the driving frequency f = fo, maximum amplitude results Tacoma Narrows Bridge 1989 freeway collapse Shattering a glass by singing

Wave Medium Mechanical Waves Electromagnetic Waves Require a medium Water waves Sound waves Medium moves up and down but wave moves sideways Electromagnetic Waves Do not require a medium EM waves can travel through the vacuum of space

Parts of a wave Crest Trough Amplitude Wavelength Frequency (cycles/s or Hertz (Hz)) Velocity v = lf

Velocity of Waves in a String Depends on: Tension (FT) [tighter string, faster wave] Mass per unit length (m/L) [heavier string, more inertia] v = FT m/L

Example 1: Strings A wave of wavelength 0.30 m is travelling down a 300 m long wire whose total mass is 15 kg. If the wire has a tension of 1000 N, what is the velocity and frequency? v = 1000 N = 140 m/s 15 kg/300 m v = lf f = v/l = 140 m/s/0.30 m = 470 Hz

Transverse and Longitudinal Waves Transverse Wave – Medium vibrates perpendicular to the direction of wave EM waves Water waves Guitar String Longitudinal Wave – Medium vibrates in the same direction as the wave Sound

Longitudinal Waves: Velocity Wave moving along a long solid rod Wire Train track vlong= E Elastic modulus r Wave moving through a liquid or gas vlong = B Bulk modulus

Ex. 1: Longitudinal Waves: Velocity How fast would the sound of a train travel down a steel track? How long would it take the sound to travel 1.0 km? vlong= E = (2.0 X 1011/7800 kg/m3)1/2 r vlong = 5100 m/s (much fast than in air) v = x/t t = x/v = 1000m/5100m/s = 0.20 s

Earthquakes Both Transverse and Longitudinal waves are produced S(Shear) –Transverse P(Pressure) – Longitudinal In a fluid, only p waves pass Center of earth is liquid iron

Energy Transported by Waves Intensity = Power transported across a unit area perpendicular to the wave’s direction I = Power = P Area 4pr2 Comparing two distances: I1r12 = I2r22

Intensity: Example 1 The intensity of an earthquake wave is 1.0 X 106 W/m2 at a distance of 100 km from the source. What is the intensity 400 km from the source? I1r12 = I2r22 I2 = I1r12/r22 I2 = (1.0 X 106 W/m2)(100 km)2/(400 km)2 I2 = 6.2 X 104 W/m2

Reflection of a Wave Hard boundary inverts the wave Exerts an equal and opposite force Loose rope returns in same direction

Continue in same direction if using another rope boundary

Constructive and Destructive Interference Destructive Constructive Interference Interference

Constructive and Destructive Interference : Phases Waves “in phase” “out of phase” in between

Resonance Standing Wave – a wave that doesn’t appear to move Node – Point of destructive interference Antinode – Point of constructive interference (think “Antinode,Amplitude) “Standing waves are produced only at the natural (resonant) frequencies.”

Resonance: Harmonics Fundamental Lowest possible frequency “first harmonic” L = ½ l First overtone (Second Harmonic) Second overtone (Third Harmonic)

Resonance: Equations L = nln n = 1, 2, 3….. 2 f = nv = nf1 2L v = lf v = FT m/L

Example 1: Resonance A piano string is 1.10 m long and has a mass of 9.00 g. How much tension must the string be under to vibrate at 131 Hz (fund. freq.)? L = nln 2 l1 = 2L = 2.20 m 1 v = lf = (2.20 m)(131 Hz) = 288 m/s

v = FT m/L v2 = FT FT = v2m = (288 m/s)2(0.009 kg) = 676 N L (1.10 m)

What are the frequencies of the first four harmonics of this string? f1 = 131 Hz 1st Harmonic f2 = 262 Hz 2nd Harmonic 1st Overtone f3 = 393 Hz 3rd Harmonic 2nd Overtone f4 = 524 Hz 4th Harmonic 3rd Overtone

Hitting a Boundary Both reflection and refraction occur Angle of incidence = angle of reflection q1 q2 q1 = q2 Reflected wave air water Refracted wave

Refraction Velocity of a wave changes when crossing between substances Soldiers slow down marching into mud sin q1 = v1 sin q2 = v2

Example 1: Refraction An earthquake p-wave crosses a rock boundary where its speed changes from 6.5 km/s to 8.0 km/s. If it strikes the boundary at 30o, what is the angle of refraction? sin q1 = v1 sin 30o = 6.5 km/s sin q2 = v2 sin q2 8.0 km/s q2 = 38o

Example 2: Refraction A sound wave travels through air at 343 m/s and strikes water at an angle of 50. If the refracted angle is 21.4o, what is the speed of sound in water? (Ans: 1440 m/s)

Diffraction Note bending of wave into “shadow region”

Diffraction Bending of waves around an object Only waves diffract, not particles The smaller the obstacle, the more diffraction in the shadow region