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We know springs have potential energy PE = ½ kx 2 F by spring = -kx F on spring = + kx (Hooke’s Law) Force (by the spring) is negative because it is in the opposite direction of the displacement (restoring force) If there was no friction, this system would continue oscillating indefinitely.

We know springs have potential energy PE = ½ kx 2 F by spring = -kx F on spring = + kx

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We know springs have potential energy PE = ½ kx 2 F by spring = -kx F on spring = + kx (Hooke’s Law) Force (by the spring) is negative because it is in the opposite direction of the displacement (restoring force) - PowerPoint PPT Presentation

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Page 1: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

We know springs have potential energy

PE = ½ kx2

Fby spring = -kx

Fon spring = + kx

(Hooke’s Law)

Force (by the spring) is negative because it is in the opposite direction of the displacement (restoring force)

If there was no friction, this system would continue oscillating indefinitely.

It would demonstrate simple harmonic motion.

Page 2: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

SHM Terms and Concepts

Amplitude (A) – the maximum displacement away from the equilibrium point (m)

Cycle – one complete motion forward and back

Period (T) – the time it takes to complete one complete cycle (s)

Frequency (f) – the number of cycles per second (Hz, s-1)

Page 3: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

A family of four with a total mass of 200 kg step into their car causing the springs to compress 3.0 cm. What is the spring constant of the springs assuming they act as a single spring? How far would the car lower if loaded with 300 kg rather than 200 kg?

Page 4: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

A spring stretches 0.150 m when a 0.300-kg mass is gently lowered on it. If the system is then placed on a frictionless table and the mass is pulled 0.100 m from its equilibrium position, determine (a) the spring constant, (b) the maximum speed attained during an oscillation, (c) the speed when the mass is 0.050 m from equilibrium, and (d) the maximum acceleration of the mass.

Page 5: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Relationship between UCM and SHM

http://www2.biglobe.ne.jp/~norimari/science/JavaEd/e-wave1.html

SHM is related to UCM in that it is UCM observed from a different perspective.

SHM is also related to wave motion

Page 6: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Oscillating Springs

The period (T) of an object in SHM is the time it takes to complete a cycle.

What variables of this frictionless system would affect the period?

Page 7: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Since energy is conserved in SHM, PEmax = KEmax

½kA2 = ½mvmax2

A2 / vmax2 = m / k

A / vmax = √(m / k)

Period of oscillating springs affected by mass and spring constant…NOT the amplitude.

The period in SHM is analogous to the period in UCMvmax = 2πA / T

A / vmax = T / 2π

T = 2 π √(m / k)

Page 8: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

A spider of mass 0.30 g waits in its web of negligible mass. A slight movement causes the web to vibrate with a frequency of 15 Hz. (a) What is the value of k for the elastic web? (b) At what frequency would you expect the web to vibrate if an insect of mass 0.10 g were trapped in addition to the spider?

Page 9: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Pendulums

Two forces act on a pendulum with length L – the weight (mg) and the tension (FT)

The force that causes the pendulum to move is tangent to the arc of movement:

F = -mgsinθ

For small angles, sin θ = θ

F = -mgθ

Since θ = x/L

F = -(mg/L)x

What determines the period of a pendulum?

Page 10: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Pendulums

F = -(mg/L)x

-F/x = mg/L

At small angles, a pendulum exhibits SHM.

SHM period equation:

T = 2π√(m/k)

Hooke’s Law: F = -kx,

k = -F/x

k = mg/L

T = 2π√(m/(mg/L))

T = 2π√(L/g)Period determined by length

and grav accel

Page 11: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

A geologist uses a simple pendulum that has a length of 37.10 cm and a frequency of 0.8190 Hz at a certain location on the Earth. What is the acceleration due to gravity at this location?

Page 12: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

The length of a simple pendulum is 0.5 m, the pendulum bob has a mass of 25 g, and it is released so that the bob is 10 cm above its equilibrium position. (a) With what frequency does it vibrate? (b) What is the pendulum bob’s speed when it passes through the lowest point of the swing?

Page 13: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Wave Motion

http://www2.biglobe.ne.jp/~norimari/science/JavaEd/e-wave1.html

Objects that vibrate in SHM can produce periodic waves which share many of the same characteristics as SHM objects.

Mechanical wave – a disturbance that travels through some type of medium or substance

Seismic waves Ocean wavesSound waves

Nonmechanical wave – a disturbance that does not require a medium or substance to travel (Electromagnetic radiation aka light)

Page 14: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Transverse (shear) waves cause particles to move back and forth at right angles to the direction of wave travel

Longitudinal (compressional) waves cause particles to move back and forth in the same direction as wave travel (SOUND is this type)

Page 15: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

A few characteristics about transverse waves:

Crest – the high point of a wave (the “peak”)Trough – the low point of a wave (the “valley”)Wavelength – the distance between two successive crests (or troughs) in

a waveAmplitude – the distance from the midpoint to the crest (or the trough) of

a waveFrequency – the number of wavelengths that pass a given point in one

second Period – the amount of time it takes for one wavelength to pass a given

point

Page 16: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Sound as a Wave

Sound is produced by a vibrating object which creates a disturbance of air molecules in the form of a longitudinal wave.

Speed of sound (air) = (331 + 0.60T) m/s

T = temperature in celsius

Path of vibration

air

eardrum

three bones

cochlear window

cochlear fluid

Page 17: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

The speed of a wave is equal to the frequency times the wavelength.

V = f λ

v = speed (m/s)

f = frequency (Hz, s-1)

λ = wavelength (m)

Page 18: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

A periodic wave has a wavelength of 0.50 m and a speed of 20 m/s. What is the wave frequency?

Radio waves from an FM station have a frequency of 103.1 MHz. If the waves travel with a speed of 3.00 x 108 m/s, what is the wavelength?

Page 19: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Resonance

The frequency determined using the SHM formulas is called the natural frequency.

1 / f = 2 π √(m / k) 1 / f = 2 π √(L / g)The system vibrates at this frequency on its own. Without friction, it would vibrate indefinitely.

If an external force is applied at a given frequency, the closer it matches the natural frequency, the greater the amplitude will be.

When an externally applied frequency matches an object’s natural frequency, you have resonance.

Example:Child being pushed on swingMarching in step across bridgeBreaking glass with voice

Page 20: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Resonance in Waves

When two waves are traveling toward each other, their interaction obeys the principle of superposition.

This states that the resultant wave is the algebraic sum of their individual displacements.

Page 21: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Situation (b) is called constructive interference because the two individual waves combine to make a larger wave.

Situation (a) is called destructive interference because the two individual waves cancel each other out.

Page 22: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Constructive and destructive interference can occur simultaneously.

Where a crest meets a crest (constructive), the waves are in phase.

Where a crest meets a trough (destructive), the waves are out of phase.

Page 23: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Situation (a) – waves in phase creating constructive interference

Situation (b) – waves completely out of phase creating destructive interference

Situation (c) – waves partially out of phase creating partial destructive interference

Page 24: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

When periodic waves are created in a cord, interference occurs between initial and reflected waves.

When the frequency of waves matches the cord’s natural frequency, a standing wave is produced.

Standing waves do not appear to be traveling, but rather appear to oscillate.

They have nodes (points of destructive interference) and antinodes (points of constructive interference)

Page 25: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Fundamental (1st Harmonic)

L = (1/2)λ

f1

First Overtone (2nd Harmonic)

L = λ

f2 = 2f1

Second Overtone (3rd Harmonic)

L = (3/2)λ

f3 = 3f1

Page 26: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

The velocity of waves on a string is 92 m/s. If the frequency of standing waves is 475 Hz, how far apart are two adjacent nodes?

Page 27: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Resonance in Musical Instruments

Wind instruments produce sound when standing waves are produced inside of them

Closed Tube

Page 28: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Resonance in Musical Instruments

Wind instruments produce sound when standing waves are produced inside of them

Open Tube

Page 29: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

What resonant frequency would you expect from blowing across the top of an empty soda bottle that is 18 cm deep assuming it is a closed tube? How would it change if it was 1/3 full of soda?

Page 30: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Beats

When two frequencies of sound are produced that are similar but not identical, their interference produces beats

Page 31: We know springs have potential energy PE = ½ kx 2 F by spring  = -kx  F on spring  = + kx

Doppler Effect

When the source of a sound wave is moving, the frequency of the sound is altered based on the position of the observer.

If the source is moving toward the observer, a higher frequency is observed (higher pitch)

If the source is moving away from the observer, a lower frequency is observed (lower pitch)