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Waves and Sound

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

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wall equilibrium position frictionless floor x = 0 F elas = 0 x max – x max v = 0 x = max F elas = max 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. …a 240 g mouse, with the same spring and surface. = 3.1 s = 0.64 s

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= 0.99 N/m What stiffness must a spring have so that the period of the mouses motion is the same as that of the cat? Ballpark answer: Need a less stiff spring; k < 22.8 N/m.

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= 1.2 Hz 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 10 4 N/m. Find the frequency of the vehicles motion after it hits a pothole.

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restoring force: simple harmonic motion (SHM): For a mass-spring system, Hookes law applies: acts to move an object back to equilibrium F restore x F restore = F elas = k x As displacement increases, so does F restore. And when x = 0… F restore = 0.

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KE = ½ m v 2 energy: kinetic energy, KE: energy of mass m having velocity v m (kg); v (m/s) the ability to do work KE = max. at eq. pos.; KE = 0 at x max.

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PE elas = ½ k ( x) 2 potential energy, PE: stored energy For a spring with spring constant k and stretch x: k (N/m); x (m) For a mass m at a height h above a reference line: m (kg); h (m); g = 9.81 m/s 2 For m-s sys., PE elas is max at x max and 0 at eq. pos. PE g = m g h For pend., PE g is max at x max and min. at eq. pos.

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

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

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period of a simple pendulum: The period of a pendulum is 5.2 s. Find… A. …its length B. …the mass of the bob Period T is independent of mass and amplitude. = 6.7 m NOT ENOUGH INFORMATION

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

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transverse waves: particles of medium move to direction of wave travel For a transverse wave: amplitude A crest wavelength trough Energy Amplitude 2

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

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

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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 A1A1 A2A2 A 1 + A 2 A1A1 A2A2 A2A2 A1A1 A 1 – A 2 A2A2 A1A1

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= 1.8 m/s Wave Velocity Equation: v (m/s); (m); T (s); f (Hz) A water wave of wavelength 8.5 m washes past a boat at anchor every 4.75 seconds. Find the waves velocity.

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v sound = T a 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: T a = air temp. in o C

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

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f n = n f 1 Waves travel along a 96.1 cm guitar string at 492 m/s. Find the fundamental frequency of the string. Find the frequency of the 5 th harmonic. frequency of the n th harmonic: 1 = 2 L = 2 (0.961 m)= m v = f and v = f 1 1 = 256 Hz 5 = 2 / 5 L = m = 1280 Hz

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Standing Waves in Open Tubes wavelength of n th harmonic of an open tube: n = 1 L 1 = 2 L n = 2 2 = L n = 3 3 = 2 / 3 L (n = 1,2,3,…)

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Closed Tubes wavelength of n th harmonic of a closed tube: n = 1 L 1 = 4 L n = 2 2 = 2 L n = 3 3 = 4 / 3 L (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 20.0 o C m 1 = 2 L = 2 (1.24 m) = 2.48 m v = f and v = f 1 1 = 138 Hz v = ? v sound = T a v sound = (20) = 343 m/s

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Find fundamental frequency for a closed tube of length 1.24 m. Air temp. is 20.0 o C. 1 = 4 L = 4 (1.24 m) = 4.96 m v = f and v = f 1 1 = 69.0 Hz 1.24 m (same as prev. prob.) doubling (or halving) of frequency = 100 Hz200 Hz400 Hz800 Hz (three octaves) one octave

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Sound compression: rarefaction: 20 Hz20,000 Hz high pressure / high density low pressure / low density audible frequencies (human hearing) infrasonicultrasonic

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Fundamental frequency determines pitch. high pitch high f = low f = = short low pitch = long

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Number and intensity of an instruments harmonics give it its unique sound quality, or ________. timbre f 1 f 2 f 3 f 4

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

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Other examples of Doppler effect: police radar race cars expansion of universe dolphins (echolocation) R O Y G B V Sun most stars (red-shifted)

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Traveling Very Fast v bug = 0 v bug < v wave v bug = v wave v bug > v wave bow wave wave barrier

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supersonic: faster than sound (vs. subsonic) sonic boom: shock wave: a 3-D bow wave caused by high-pressure air, not roaring engine cracking bullets lion tamers whip The Matrix

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= W/m 2 Sound Intensity If a pianos power output is W, find the sound intensity at a distance of… A. …1.0 m B. …2.0 m = W/m 2

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Intensity is related to volume (or relative intensity): 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 dB 40 dB 60 dB 90 dB how loud we perceive a sound to be half as loud 1 / 10 as intense 8X louder 1000X more intense

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

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

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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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F restore = F elas = k x KE = ½ m v 2 PE elas = ½ k ( x) 2 PE g = m g h v sound = T a f n = n f 1

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h h

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