harmonic motion and waves. simple harmonic motion(shm) vibration (oscillation) equilibrium position...

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 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

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 = -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)

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)– 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

All PE

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

mv2 = 19.6 N/m(0.100m2 – 0.0500m2) = 0.49 m2/s2

0.300 kgv = 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) = 2A

T = 2 m k

f = 1 T

• 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

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

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

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

Cosines and Sines

• Imagine placing a pen on a vibrating mass

• Draws a cosine wave

x = A cos2t or x = A cos2ft

T

A = Amplitude

t = time

T = period

f = frequency

x = A cos2ft

v = -vosin2ft

a = -aocos2ft

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 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

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

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

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

a = k x

m

a = (2f)2x = (2f)2A

a = [(2)(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 t

8.0

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

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

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

Remember the circle (360o = 2 rad)

= x

L

Fr = mgsin

at small angles sin= Fr = mg

L

x

mg

Fr

Fr = mg

Fr = mgx (Look’s like Hook’s Law F = -kx)

L

k = mg

L

T = 2 m

k

T = 2 mL

mg

T = 2 L

g

f = 1 = 1 g

T 2 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

2 L decrease the denominator

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

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

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– 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

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 = 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 track

vlong= E Elastic modulus

• 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

vlong = 5100 m/s (much fast than in air)

v = x/t

t = x/v1000m/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 4r2

Comparing two distances:

I1r12 = I2r2

2

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

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

First overtone (Second Harmonic)

Second overtone (Third Harmonic)

Resonance: Equations

L = nn n = 1, 2, 3…..

2

f = nv = nf1

2L

v = f

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 = nn

2

1 = 2L = 2.20 m

1

v = f = (2.20 m)(131 Hz) = 288 m/s

v = FT

m/L

v2 = FT

m/L

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

air

water

1 2 1 = 2

Refracted wave

Reflected wave

Refraction•Velocity of a wave changes when crossing between substances

•Soldiers slow down marching into mud

sin 1 = v1

sin 2 = 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 1 = v1 sin 30o = 6.5 km/s

sin 2 = v2 sin 28.0 km/s

2 = 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