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Waves and Sound

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**Simple Harmonic Motion**

period, T: time to complete one cycle (s) frequency, f: # of cycles per second (Hz)

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wall equilibrium position frictionless floor Dx = 0 Felas = 0 Dxmax v = 0 Dx = max Felas = max –Dxmax period of a mass-spring system: m = mass (kg) k = spring constant (N/m)

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**A 5.5 kg cat is attached to a fixed**

horizontal spring of stiffness 22.8 N/m and is set in motion on a frictionless surface. Find the period of motion of… …the cat. = s …a 240 g mouse, with the same spring and surface. = s

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**What stiffness must a spring have so that the period**

of the mouse’s motion is the same as that of the cat? Ballpark answer: Need a “less stiff” spring; k < 22.8 N/m. = N/m

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**A 1275 kg car carries two passengers**

with a combined mass of 153 kg. The car has four shock absorbers, each with a spring constant of 2.0 x 104 N/m. Find the frequency of the vehicle’s motion after it hits a pothole. = Hz

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restoring force: acts to move an object back to equilibrium simple harmonic motion (SHM): Frestore Dx As displacement increases, so does Frestore. And when Dx = 0… Frestore = 0. For a mass-spring system, Hooke’s law applies: Frestore = Felas = k Dx

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energy: the ability to do work kinetic energy, KE: energy of mass m having velocity v KE = ½ m v2 m (kg); v (m/s) KE = max. at eq. pos.; KE = 0 at Dxmax.

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**potential energy, PE: stored energy**

For a spring with spring constant k and “stretch” Dx: PEelas = ½ k (Dx)2 k (N/m); Dx (m) For m-s sys., PEelas is max at Dxmax and 0 at eq. pos. For a mass m at a height h above a reference line: PEg = m g h m (kg); h (m); g = 9.81 m/s2 For pend., PEg is max at Dxmax and min. at eq. pos.

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**Period T is not affected by amplitude A.**

amplitude, A: maximum displacement from equilibrium frictionless A A Period T is not affected by amplitude A. Energy of a Mass-Spring System total energy Energy (J) PEelas KE PEg

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The Pendulum For Q < 15o, a simple pendulum approximates SHM. length L amplitude Q mass m of bob Energy of a Simple Pendulum total energy Energy (J) PEg KE PEelas

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**period of a simple pendulum:**

Period T is independent of mass and amplitude. The period of a pendulum is 5.2 s. Find… A. …its length = m B. …the mass of the bob NOT ENOUGH INFORMATION

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Waves vibrations moving through space and time Waves transmit energy, not matter. medium: the matter through which the energy of mechanical waves moves

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**transverse waves: particles of medium move**

to direction of wave travel crest amplitude A trough wavelength l For a transverse wave: Energy Amplitude2

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particles of medium move // to direction of wave travel longitudinal (compressional) wave: compression rarefaction l pulse wave: a single vibration periodic wave: rhythmic, repeated vibrations

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Wave Reflection fixed boundary free boundary A A v v v v waves are reflected and inverted waves are reflected and upright

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**Two waves (unlike two objects) can occupy **

Wave Interference Two waves (unlike two objects) can occupy the same place at the same time. This condition is called interference. constructive interference: destructive interference: displacements are in same direction displacements are in opposite directions A1 A2 A1 A2 A1 + A2 A1 – A2 A2 A1 A1 A2

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Wave Velocity v (m/s); l (m); T (s); f (Hz) Equation: A water wave of wavelength 8.5 m washes past a boat at anchor every 4.75 seconds. Find the wave’s velocity. = m/s

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**The velocity of any mechanical wave **

depends only on properties of medium through which it travels. e.g., string tension, water depth, air temperature, material density, type of material An empirical equation for the velocity of sound in air: Ta = air temp. in oC vsound = Ta

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Standing Waves incident and reflected waves interfere so that antinodes have a max. amplitude, while nodes have zero amplitude On a string, nodes remain motionless; antinodes go from max. (+) to max. (–) displacement.

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**L l1 = 2 L l2 = L l3 = 2/3 L l4 = ½ L wavelength of nth**

n = 1; 1st harmonic l1 = 2 L (fundamental) n = 2; 2nd harmonic l2 = L (1st overtone) n = 3; 3rd harmonic l3 = 2/3 L (2nd overtone) n = 4; 4th harmonic l4 = ½ L (3rd overtone) wavelength of nth harmonic on a string: (n = 1,2,3,…)

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Waves travel along a 96.1 cm guitar string at 492 m/s. Find the fundamental frequency of the string. l1 = 2 L = 2 (0.961 m) = m v = f l and v = f1 l1 = Hz Find the frequency of the 5th harmonic. l5 = 2/5 L = m = Hz frequency of the nth harmonic: fn = n f1

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**Standing Waves in Open Tubes**

L n = 1 n = 2 n = 3 l1 = 2 L l2 = L l3 = 2/3 L wavelength of nth harmonic of an open tube: (n = 1,2,3,…)

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Closed Tubes L n = 1 n = 2 n = 3 l1 = 4 L l2 = 2 L l3 = 4/3 L wavelength of nth harmonic of a closed tube: (n = 1,3,5,…) (even harmonics are not present)

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**Find the fundamental frequency**

for an open tube of length 1.24 m. Assume the air temperature to be 20.0oC. 1.24 m l1 = 2 L = 2 (1.24 m) = 2.48 m v = f l and v = f1 l1 v = ? vsound = Ta vsound = (20) = 343 m/s = Hz

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**Find fundamental frequency for**

a closed tube of length 1.24 m. Air temp. is 20.0oC. 1.24 m l1 = 4 L = 4 (1.24 m) = 4.96 m v = f l and v = f1 l1 (same as prev. prob.) = Hz doubling (or halving) of frequency = one octave 100 Hz 200 Hz 400 Hz 800 Hz (three octaves)

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Sound compression: high pressure / high density rarefaction: low pressure / low density audible frequencies (human hearing) 20 Hz 20,000 Hz infrasonic ultrasonic

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**Fundamental frequency determines pitch.**

high f = high pitch = short l low f = low pitch = long l

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**Number and intensity of an instrument’s harmonics **

give it its unique sound quality, or ________. timbre f1 f2 f3 f4 f1 f2 f3 f4 f1 f2 f3 f4 f1 f2 f3 f4

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**Relative motion between wave source and observer **

The Doppler Effect Relative motion between wave source and observer causes a change in the ____________ frequency. observed v = 0 femitted fobserved (higher) femitted femitted fobserved (lower)

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**Other examples of Doppler effect:**

race cars police radar R O Y G B V Sun most stars (“red-shifted”) dolphins (echolocation) expansion of universe

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Traveling Very Fast vbug = 0 vbug < vwave wave barrier bow wave vbug > vwave vbug = vwave

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**supersonic: “faster than sound” (vs. subsonic)**

shock wave: a 3-D bow wave sonic boom: caused by high-pressure air, not roaring engine lion tamer’s whip cracking bullets The Matrix

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Sound Intensity If a piano’s power output is W, find the sound intensity at a distance of… A. …1.0 m = W/m2 B. …2.0 m = W/m2

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**Intensity is related to volume (or relative intensity):**

-- how loud we perceive a sound to be -- measured in decibels (dB) A difference of 10 dB changes the sound intensity by a factor of 10 and the volume by a factor of 2. 50 dB 40 dB 60 dB 90 dB half as loud 8X louder 1/10 as intense 1000X more intense

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**alternating loud-and-soft sounds resulting **

Beats alternating loud-and-soft sounds resulting from interference between two slightly- different frequencies Equation: 1 second elapses 1 second elapses f1 = 16 Hz f1 = 16 Hz f2 = 17 Hz f2 = 18 Hz fbeat = 1 Hz fbeat = 2 Hz

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**Forced Vibrations and Resonance**

natural frequency: the frequency at which an object most easily vibrates forced vibration: a vibration due to an applied force resonance: occurs when a force is repeatedly applied to an object AT the object’s natural frequency -- result of resonance = large amplitude

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**shattering crystal wine glasses**

Examples: swing shattering crystal wine glasses Tacoma Narrows Bridge (1940) British regiment (Manchester, 1831) aeolian harps “The wind in the wires made a tattletale sound, as a wave broke over the railing…”

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vsound = Ta Frestore = Felas = k Dx KE = ½ m v2 PEelas = ½ k (Dx)2 fn = n f1 PEg = m g h

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h h

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Phys 250 Ch15 p1 Chapter 15: Waves and Sound Example: pulse on a string speed of pulse = wave speed = v depends upon tension T and inertia (mass per length.

Phys 250 Ch15 p1 Chapter 15: Waves and Sound Example: pulse on a string speed of pulse = wave speed = v depends upon tension T and inertia (mass per length.

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