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Harmonic Motion and Waves

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

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

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Period and Frequency A radio station has a frequency of M Hz. What is the period of the wave? M Hz1X10 6 Hz= X 10 8 Hz 1M Hz T = 1/f = 1/(1.031 X 10 8 Hz) = X s

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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

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

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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)

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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/s 2 )/(-0.03 m) k = 6.5 X 10 4 N/m (note that we did not include the mass of the car)

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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/s 2 )/ 6.5 X 10 4 N/m x = 4.5 X m = 4.5 cm

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

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Energy and Springs KE = ½ mv 2 PE = ½ kx 2 Maximum PE = ½ kA 2 Law of conservation of Energy ½ kA 2 = ½ mv 2 + ½ kx 2

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All PE All KE All PE Some KE and Some PE

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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 kA 2 Maximum PE = ½ (20 N/m)(0.030 m 2 ) Maximum PE = 9 X Nm (J)

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Spring Energy: Example 1a What is the maximum speed of the mass? ½ kA 2 = ½ mv 2 + ½ kx 2 ½ kA 2 = ½ mv 2 (x=0 at the origin) 9 X J = ½ (0.50 kg)v 2 v = 0.19 m/s

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Spring Energy: Example 1b What is the potential energy and kinetic energy at x = 2.0 cm? PE = ½ kx 2 PE = ½ (20 N/m)(0.020 m 2 ) = 4 X J ½ kA 2 = ½ mv 2 + ½ kx 2 ½ mv 2 = ½ kA 2 - ½ kx 2 KE = 9 X J - 4 X J = 5 X J

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

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

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

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The maximum velocity occurs through the origin: ½ kA 2 = ½ mv 2 + ½ kx 2 ½ kA 2 = ½ mv 2 (x=0 at the origin) kA 2 = mv 2 v 2 = kA 2 /m v = \/kA 2 /m = \/(19.6 N/m)(0.100m) 2 /0.300kg v = m/s

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Spring Energy: Example 2c What is the velocity at x = m? ½ kA 2 = ½ mv 2 + ½ kx 2 kA 2 = mv 2 + kx 2 mv 2 = kA 2 - kx 2 v 2 = kA 2 - kx 2 m v 2 = 19.6 N/m(0.100m 2 – m 2 ) = 0.49 m 2 /s kg v = m/s

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

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

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Period depends only on mass and spring constant Amplitude does not affect period v o = 2 Aforv o = 2 A T v o is the initial (and maximum) velocity

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Period: Example 1 What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 10 4 N/m after it hits a bump? T = 2 m= 2 (1400 kg/6.5 X 10 4 N/m) 1/2 k T = 0.92 s f = 1/T = 1/0.92 s = 1.09 Hz

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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 = s T = 2 m k T 2 = (2 ) 2 m k k = (2 ) 2 m = (2 ) 2 (3.0 X kg) = 2.7 N/m T 2 (0.0667) 2

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

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

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x = A cos2 torx = A cos2 ft T A = Amplitude t = time T = period f = frequency

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x = A cos2 ft v = -v o sin2 ft a = -a o cos2 ft Velocity is the derivative of position Acceleration is the derivative of velocity

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Cos: Example 1a A loudspeaker vibrates at 262 Hz (middle C). The amplitude of the cone of the speaker is 1.5 X m. What is the equation to describe the position of the cone over time? x = A cos2 ft x = (1.5 X m) cos2 (262 s -1 )t x = (1.5 X m) cos(1650 s -1 )t

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Cos: Example 1b What is the position at t = 1.00 ms (1 X s) x = A cos2 ft x = (1.5 X m) cos2 (262 s -1 ) (1 X s) x = (1.5 X m) cos(1.65 rad) = -1.2 X m

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Cos: Example 1c What is the maximum velocity and acceleration? v o = 2 Af v o = 2 (1.5 X m)(262 s -1 ) = 0.25 m/s

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F = ma kx = ma a = kx/mBut we don’t know k or m a = k xSolve for k/m m T = 2 m k T 2 = (2 ) 2 m k k = (2 ) 2 = (2 ) 2 f 2 mT2mT2

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a = k x m a = (2 f) 2 x = (2 f) 2 A a = [(2 )(262 Hz)] 2 (1.5 X m) = 410 m/s 2

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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 t 8.0

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x = A cos2 ft x = (0.25 m)cos t 8.0 Therefore A = 0.25 m 2 ft = t 8.0 2f = 8.0 f = 1/16 HzT = 1/f = 16 s

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

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

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Remember the circle (360 o = 2 rad) = x L F r = mgsin at small angles sin = F r = mg L x mg FrFr

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

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

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

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

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

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

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

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Wave Medium Mechanical 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

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

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

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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 = ff = v/ = 140 m/s/0.30 m = 470 Hz

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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

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Longitudinal Waves: Velocity Wave moving along a long solid rod –Wire –Train track v long = EElastic modulus Wave moving through a liquid or gas v long = BBulk modulus

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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? v long = E = (2.0 X /7800 kg/m 3 ) 1/2 v long = 5100 m/s (much fast than in air) v = x/t t = x/v 1000m/5100m/s = 0.20 s

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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

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Energy Transported by Waves Intensity = Power transported across a unit area perpendicular to the wave’s direction I = Power = P Area 4 r 2 Comparing two distances: I 1 r 1 2 = I 2 r 2 2

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Intensity: Example 1 The intensity of an earthquake wave is 1.0 X 10 6 W/m 2 at a distance of 100 km from the source. What is the intensity 400 km from the source? I 1 r 1 2 = I 2 r 2 2 I 2 = I 1 r 1 2 /r 2 2 I 2 = (1.0 X 10 6 W/m 2 )(100 km) 2 /(400 km) 2 I 2 = 6.2 X 10 4 W/m 2

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

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Continue in same direction if using another rope boundary

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Constructive and Destructive Interference DestructiveConstructiveInterference

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

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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.”

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

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Resonance: Equations L = n n n = 1, 2, 3….. 2 f = nv = nf 1 2L v = f v = F T m/L

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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 = n n 2 1 = 2L = 2.20 m 1 v = f = (2.20 m)(131 Hz) = 288 m/s

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v = F T m/L v 2 = F T m/L F T = v 2 m= (288 m/s) 2 (0.009 kg) = 676 N L(1.10 m)

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What are the frequencies of the first four harmonics of this string? f 1 = 131 Hz1 st Harmonic f 2 = 262 Hz2 nd Harmonic1 st Overtone f 3 = 393 Hz3 rd Harmonic2 nd Overtone f 4 = 524 Hz4 th Harmonic3 rd Overtone

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Hitting a Boundary Both reflection and refraction occur Angle of incidence = angle of reflection air water 11 2 1 = 2 Refracted wave Reflected wave

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Refraction Velocity of a wave changes when crossing between substances Soldiers slow down marching into mud sin 1 = v 1 sin 2 = v 2

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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 30 o, what is the angle of refraction? sin 1 = v 1 sin 30 o = 6.5 km/s sin 2 = v 2 sin km/s 2 = 38 o

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Example 2: Refraction A sound wave travels through air at 343 m/s and strikes water at an angle of 5 0. If the refracted angle is 21.4 o, what is the speed of sound in water? (Ans: 1440 m/s)

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Diffraction Note bending of wave into “shadow region”

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

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