# Simple Harmonic Motion(SHM) Harmonic Motion and Waves. Period and Frequency. Period and Frequency

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1 Simple Harmonic Motion(SHM) Harmonic Motion and Waves 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 (bac to starting point) Frequency (Hz) Cycles per second F = 1 T = 1 T f Period and Frequency A radio station has a frequency of M Hz. What is the period of the wave? M Hz 1X10 6 Hz = X 10 8 Hz 1M Hz T = 1/f = 1/(1.031 X 10 8 Hz) = X 10-9 s 1

2 F = -x Hooe s Law F = weight of an object = spring constant (N/m) x = displacement when the object is placed on the spring Hooe s Law: Example 1 What is the spring constant if a g 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) Hooe s Law: Example A family of four has a combined mass of 00 g. When they step in their 100 g car, the shocs compress 3.0 cm. What is the spring constant of the shocs? F = -x = -F/x = -(00 g)(9.8 m/s )/(-0.03 m) = 6.5 X 10 4 N/m (note that we did not include the mass of the car) Hooe s Law: Example a How far will the car lower if a 300 g family borrows the car? F = -x x = -F/ x = -(300 g)(9.8 m/s )/ 6.5 X 10 4 N/m x = 4.5 X 10 - m = 4.5 cm Forces on a Spring Extreme Position (Amplitude) Force at maximum Velocity = 0 Equilibrium position Force = 0 Velocity at maximum

3 Energy and Springs KE = ½ mv PE = ½ x Maximum PE = ½ A Law of conservation of Energy All PE All KE All PE ½ A = ½ mv + ½ x Some KE and Some PE Spring Energy: Example 1 A 0.50 g mass is connected to a light spring with a spring constant of 0 N/m. Calculate the total energy if the amplitude is 3.0 cm. Maximum PE = ½ A Maximum PE = ½ (0 N/m)(0.030 m ) Maximum PE = 9 X 10-3 Nm (J) Spring Energy: Example 1a What is the maximum speed of the mass? ½ A = ½ mv + ½ x ½ A = ½ mv 9 X 10-3 J = ½ (0.50 g)v v = 0.19 m/s (x=0 at the origin) Spring Energy: Example 1b What is the potential energy and inetic energy at x =.0 cm? Spring Energy: Example 1c At what position is the speed 0.10 m/s? PE = ½ x PE = ½ (0 N/m)(0.00 m ) = 4 X 10-3 J ½ A = ½ mv + ½ x ½ mv = ½ A - ½ x KE = 9 X 10-3 J - 4 X 10-3 J = 5 X 10-3 J (Ans: +.6 cm) 3

4 Spring Energy: Example a A spring stretches m when a g mass is suspended from it (diagrams a and b). Find the spring constant. Spring Energy: Example b The spring is now stretched an additional m and allowed to oscillate (diagram c). What is the maximum velocity? (Ans: 19.6 N/m) The maximum velocity occurs through the origin: ½ A = ½ mv + ½ x ½ A = ½ mv (x=0 at the origin) A = mv v = A /m v = \/A /m = \/(19.6 N/m)(0.100m) /0.300g v = m/s Spring Energy: Example c What is the velocity at x = m? ½ A = ½ mv + ½ x A = mv + x mv = A - x v = A - x m v = 19.6 N/m(0.100m m ) = 0.49 m /s g v = m/s Spring Energy: Example d What is the maximum acceleration? The force is a maximum at the amplitude F = ma and F = x ma = x a = x/m = (19.6 N/m)(0.100 m)/(0.300 g) a = 6.53 m/s Trigonometry and SHM Ball rotates on a table Loos lie a spring from the side One rev(diameter) = A T = m f = 1 T 4

5 Period depends only on mass and spring constant Amplitude does not affect period v o = Af or v o = A T v o is the initial (and maximum) velocity Period: Example 1 What is the period and frequency of a 1400 g car whose shocs have a of 6.5 X 10 4 N/m after it hits a bump? T = m = (1400 g/6.5 X 10 4 N/m) 1/ T = 0.9 s f = 1/T = 1/0.9 s = 1.09 Hz Period: Example a 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 = m T = ( ) m = ( ) m = ( ) (3.0 X 10-4 g) =.7 N/m T (0.0667) Period: Example b What would be the frequency for a lighter insect, 0.10 g? Would it be higher or lower? T = m T = (m/) 1/ T = (1.0 X 10-4 g/.7 N/m) 1/ = s f = 1/T = 1/0.038 s = 6 Hz Cosines and Sines Imagine placing a pen on a vibrating mass Draws a cosine wave x = A cos t or x = A cos ft T A = Amplitude t = time T = period f = frequency 5

6 x = A cos ft Velocity is the derivative of position v = -v o sin ft a = -a o cos ft Acceleration is the derivative of velocity Cos: Example 1a A loudspeaer vibrates at 6 Hz (middle C). The amplitude of the cone of the speaer is 1.5 X 10-4 m. What is the equation to describe the position of the cone over time? x = A cos ft x = (1.5 X 10-4 m) cos (6 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 cos ft x = (1.5 X 10-4 m) cos (6 s -1 ) (1 X 10-3 s) x = (1.5 X 10-4 m) cos(1.65 rad) = -1. X 10-5 m Cos: Example 1c What is the maximum velocity and acceleration? v o = Af v o = (1.5 X 10-4 m)(6 s -1 ) = 0.5 m/s F = ma x = ma a = x/m But we don t now or m a = x Solve for /m m T = m T = ( ) m = ( ) = ( ) f m T a = x m a = ( f) x = ( f) A a = [( )(6 Hz)] (1.5 X 10-4 m) = 410 m/s 6

7 Cos: Example a Find the amplitude, frequency and period of motion for an object vibrating at the end of a spring that follows the equation: x = A cos ft x = (0.5 m)cos t 8.0 Therefore A = 0.5 m x = (0.5 m)cos 8.0 t ft = t 8.0 f = 8.0 f = 1/16 Hz T = 1/f = 16 s Cos: Example b Find the position of the object after.0 seconds. x = (0.5 m)cos t 8.0 x = (0.5 m)cos 4.0 x = 0.18 m 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 Remember the circle (360 o = rad) x mg L = x L F r = mgsin at small angles sin = F r = mg F r = mg F r = mgx L = mg L T = m T = ml mg (Loo s lie Hoo s Law F = -x) F r 7

8 T = L g f = 1 = 1 g T 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 (tucing and extending your legs) f = 1 g L decrease the denominator Example 1: Pendulum What would be the period of a grandfather cloc with a 1.0 m long pendulum? T = L g Ans:.0 s Example : Pendulum Estimate the length of the pendulum of a grandfather cloc that tics once per second (T = 1.0 s). T = L g Ans: 0.5 m Damped Harmonic Motion Most SHM systems slowly stop For car shocs, a fluid dampens the motion 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 8

9 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 Parts of a wave Crest Trough Amplitude Wavelength Frequency (cycles/s or Hertz (Hz)) Velocity v = f 9

10 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 Example 1: Strings A wave of wavelength 0.30 m is travelling down a 300 m long wire whose total mass is 15 g. If the wire has a tension of 1000 N, what is the velocity and frequency? v = 1000 N = 140 m/s 15 g/300 m v = f f = v/ = 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 trac v long = E Elastic modulus Wave moving through a liquid or gas v long = B Bul modulus 10

11 Ex. 1: Longitudinal Waves: Velocity How fast would the sound of a train travel down a steel trac? How long would it tae the sound to travel 1.0 m? v long = E = (.0 X /7800 g/m 3 ) 1/ v long = 5100 m/s (much fast than in air) v = x/t t = x/v 1000m/5100m/s = 0.0 s Earthquaes 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 4 r Comparing two distances: I 1 r 1 = I r Intensity: Example 1 The intensity of an earthquae wave is 1.0 X 10 6 W/m at a distance of 100 m from the source. What is the intensity 400 m from the source? I 1 r 1 = I r I = I 1 r 1 /r I = (1.0 X 10 6 W/m )(100 m) /(400 m) I = 6. X 10 4 W/m Reflection of a Wave Hard boundary inverts the wave Exerts an equal and opposite force Continue in same direction if using another rope boundary Loose rope returns in same direction 11

12 Constructive and Destructive Interference Constructive and Destructive Interference : Phases Destructive Interference Constructive Interference Waves in phase out of phase in between Resonance Resonance: Harmonics Standing Wave a wave that doesn t appear to move Node Point of destructive interference Antinode Point of constructive interference (thin Antinode,Amplitude) Fundamental Lowest possible frequency first harmonic L = ½ First overtone (Second Harmonic) Second overtone (Third Harmonic) Standing waves are produced only at the natural (resonant) frequencies. L = n n Resonance: Equations n = 1,, 3.. 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.)? f = nv = nf 1 L v = f v = F T m/l L = n n 1 = L =.0 m 1 v = f = (.0 m)(131 Hz) = 88 m/s 1

13 v = F T m/l v = F T m/l F T = v m = (88 m/s) (0.009 g) = 676 N L (1.10 m) What are the frequencies of the first four harmonics of this string? f 1 = 131 Hz 1 st Harmonic f = 6 Hz nd Harmonic 1 st Overtone f 3 = 393 Hz 3 rd Harmonic nd Overtone f 4 = 54 Hz 4 th Harmonic 3 rd Overtone Hitting a Boundary Both reflection and refraction occur Angle of incidence = angle of reflection 1 1 = Refraction Velocity of a wave changes when crossing between substances Soldiers slow down marching into mud sin 1 = v 1 sin = v Reflected wave Refracted wave air water Example 1: Refraction An earthquae p-wave crosses a roc boundary where its speed changes from 6.5 m/s to 8.0 m/s. If it stries the boundary at 30 o, what is the angle of refraction? Example : Refraction A sound wave travels through air at 343 m/s and stries water at an angle of 5 0. If the refracted angle is 1.4 o, what is the speed of sound in water? sin 1 = v 1 sin 30 o = 6.5 m/s sin = v sin 8.0 m/s = 38 o (Ans: 1440 m/s) 13

14 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 14

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