Harmonic Motion and Waves. Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position...
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Transcript of Harmonic Motion and Waves. Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position...
![Page 1: Harmonic Motion and Waves. Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude.](https://reader036.fdocuments.in/reader036/viewer/2022062320/56649c9d5503460f9495c2b7/html5/thumbnails/1.jpg)
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 = 1 T = 1
T f
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Period and FrequencyA 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
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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 0.100 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 = -kxk = -F/xk = -(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)
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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/s2)/ 6.5 X 104 N/m
x = 4.5 X 10-2 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 mv2
• PE = ½ kx2
• Maximum PE = ½ kA2
Law of conservation of Energy
½ kA2 = ½ mv2+ ½ kx2
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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 kA2
Maximum PE = ½ (20 N/m)(0.030 m2)
Maximum PE = 9 X 10-3 Nm (J)
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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
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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
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Spring Energy: Example 1c
At what position is the speed 0.10 m/s?
(Ans: + 2.6 cm)
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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)
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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?
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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
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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
mv2 = 19.6 N/m(0.100m2 – 0.0500m2) = 0.49 m2/s2
0.300 kgv = 0.700 m/s
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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
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Trigonometry and SHM
• Ball rotates on a table• Looks like a spring from the side• One rev(diameter) = 2A
T = 2 m k
f = 1 T
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• Period depends only on mass and spring constant
• Amplitude does not affect period
vo = 2Af or vo = 2A
T
vo 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 104 N/m after it hits a bump?
T = 2 m = 2 (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
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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 = 0.0667 sT = 2 m
kT2 = (2)2m
kk = (2)2m = (2)2(3.0 X 10-4 kg) = 2.7 N/m
T2 (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 10-4 kg/2.7 N/m)1/2 = 0.038 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 cos2t or x = A cos2ft
T
A = Amplitude
t = time
T = period
f = frequency
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x = A cos2ft
v = -vosin2ft
a = -aocos2ft
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 10-4 m. What is the equation to describe the position of the cone over time?
x = A cos2ft
x = (1.5 X 10-4 m) cos2(262 s-1)t
x = (1.5 X 10-4 m) cos(1650 s-1)t
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Cos: Example 1b
What is the position at t = 1.00 ms (1 X 10-3 s)
x = A cos2ft
x = (1.5 X 10-4 m) cos2(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
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Cos: Example 1c
What is the maximum velocity and acceleration?
vo = 2Af
vo = 2(1.5 X 10-4 m)(262 s-1) = 0.25 m/s
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F = makx = maa = kx/m But we don’t know k or ma = k x Solve for k/m mT = 2 m
kT2 = (2)2m
kk = (2)2 = (2)2f2
m T2
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a = k x
m
a = (2f)2x = (2f)2A
a = [(2)(262 Hz)]2(1.5 X 10-4 m) = 410 m/s2
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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 cos2ftx = (0.25 m)cos t
8.0
Therefore A = 0.25 m
2ft = t 8.0
2f = 8.0
f = 1/16 Hz T = 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 (<15o)
• The restoring force is at a maximum at the top of the swing.
Fr = restoring Force
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Remember the circle (360o = 2 rad)
= x
L
Fr = mgsin
at small angles sin= Fr = mg
L
x
mg
Fr
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Fr = mg
Fr = 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 L decrease 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 (fo)
• When the driving frequency f = fo, 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 (FT) [tighter string, faster wave]
• Mass per unit length (m/L) [heavier string, more inertia]
v = FT
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 = f f = 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
vlong= E Elastic modulus
• Wave moving through a liquid or gas
vlong = B Bulk 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?
vlong= E = (2.0 X 1011/7800 kg/m3)1/2
vlong = 5100 m/s (much fast than in air)
v = x/t
t = x/v1000m/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 4r2
Comparing two distances:
I1r12 = I2r2
2
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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 = I2r2
2
I2 = I1r12/r2
2
I2 = (1.0 X 106 W/m2)(100 km)2/(400 km)2
I2 = 6.2 X 104 W/m2
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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
Destructive Constructive
Interference Interference
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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 = nn n = 1, 2, 3…..
2
f = nv = nf1
2L
v = f
v = FT
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 = nn
2
1 = 2L = 2.20 m
1
v = f = (2.20 m)(131 Hz) = 288 m/s
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v = FT
m/L
v2 = FT
m/L
FT = v2m = (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?
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
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Hitting a Boundary
• Both reflection and refraction occur
• Angle of incidence = angle of reflection
air
water
1 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 = v1
sin 2 = v2
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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 30o, what is the angle of refraction?
sin 1 = v1 sin 30o = 6.5 km/s
sin 2 = v2 sin 28.0 km/s
2 = 38o
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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)
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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