General Physics (PHY 2130)

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1 General Physics (PHY 2130) Lecture 28 Waves standing waves Sound definitions standing sound waves and instruments Doppler s effect

2 Lightning Review Last lecture: 1. Waves speed of the wave mathematical description of the wave Review Problem: By shaking one end of a stretched string, a single pulse is generated. The traveling pulse carries 1. energy 2. momentum 3. energy and momentum 4. neither of the two

3 3 Standing Waves Pluck a stretched string such that y(x,t) = A sin(ωt + kx) When the wave strikes the wall, there will be a reflected wave that travels back along the string. The reflected wave will be 180 out of phase with the wave incident on the wall. Its form is y(x,t) = -A sin (ωt - kx). The interference of the incident and reflected wave would result a standing wave: A node (N) is a point of zero oscillation. An antinode (A) is a point of maximum displacement. All points between nodes oscillate up and down.

4 The reflected wave will be 180 out of phase with the wave incident on the wall. Its form is y(x,t) = -A sin (ωt - kx). Apply the superposition principle to the two waves on the string: y( x, t) = y1( x, t) + y2 = = ( x, t) A( sin( ωt + kx) sin( ωt kx) ) ( 2Acosωt) sin kx 4 This is the mathematical form of a standing wave. The nodes occur where y(x,t) = 0. That is when sin kx = 0 The antinodes occur when y(x,t) = 2Acosωt that is when sin kx=± 1;

5 5 Consider the mathematical form of a standing wave. The nodes occur where y(x,t) = 0. y ( 2Acos t) sin kx y( x, t) = ω ( x, t) = 2Acosωt sin kx = 0 The nodes are found from the locations where sin kx=0, which happens when kx = 0, π, 2π,. That is when kx = nπ where n = 0,1,2, The antinodes occur when sin kx=± 1; that is where kx kx π 3π =,, 2 2 ( 2n + 1) π = and n 2 = 0, 1, 2,

6 Standing Waves on a String fixed at both ends 6 If the string has a length L, and both ends are fixed, then y(x=0,t) = 0 and y(x=l, t) = 0. y y ( x = 0, t) sin k( 0) ( x = L, t) sin kl = = 0 0 kl = nπ 2π L = nπ λ 2L λ = n where n = 1, 2, 3,

7 7 λ n = 2L n These are the permitted wavelengths of standing waves on a string; no others are allowed. The speed of the wave is: v = λ n f n The allowed frequencies are then: f n v nv = = n =1, 2, 3, λ 2L n

8 8 The n=1 frequency is called the fundamental frequency. f v = λ nv 2L v n 2L = n = = n nf 1 All allowed frequencies (called harmonics) are integer multiples of f 1.

9 Example: A Guitar s E-string has a length 65 cm and is stretched to a tension of 82 N. It vibrates with a fundamental frequency of Hz. Determine the mass per unit length of the string. 9 For a wave on a string: v = F µ Solving for the linear mass density: µ = = F = F = F v ( λ1 f1) f1 ( 2L) ( 82 N) 2 ( Hz) ( 2*0.65 m) 2 = kg/m

10 10 Sound Waves Sound waves are longitudinal. They can be represented by either variations in pressure (gauge pressure) or by displacements of an air element. Speed of sound in air at 0 0 C ~ 331 m/s Frequencies Audible range 20 Hz to 20 khz Infrasound - below 20 Hz Ultrasound - above 20 khz

11 11 The middle of a compression (rarefaction) corresponds to a pressure maximum (minimum).

12 Using a Tuning Fork to Produce a Sound Wave A tuning fork will produce a pure musical note As the tines vibrate, they disturb the air near them As the tine swings to the right, it forces the air molecules near it closer together This produces a high density area in the air This is an area of compression As the tine moves toward the left, the air molecules to the right of the tine spread out This produces an area of low density This area is called a rarefaction

13 Using a Tuning Fork As the tuning fork continues to vibrate, a succession of compressions and rarefactions spread out from the fork A sinusoidal curve can be used to represent the longitudinal wave Crests correspond to compressions and troughs to rarefactions

14 14 Sound Waves Sound waves are longitudinal. They can be represented by either variations in pressure (gauge pressure) or by displacements of an air element. Speed of sound in air at 0 0 C ~ 331 m/s Frequencies Audible range 20 Hz to 20 khz Infrasound - below 20 Hz Ultrasound - above 20 khz

15 Applications of Ultrasound Can be used to produce images of small objects Widely used as a diagnostic and treatment tool in medicine Ultrasonic flow meter to measure blood flow May use piezoelectric devices that transform electrical energy into mechanical energy Reversible: mechanical to electrical Ultrasounds to observe babies in the womb Cavitron Ultrasonic Surgical Aspirator (CUSA) used to surgically remove brain tumors Ultrasonic ranging unit for cameras

16 Speed of Sound v = elastic inertial property property The speed of sound is higher in solids than in gases The molecules in a solid interact more strongly The speed is slower in liquids than in solids Liquids are more compressible

17 Speed of Sound in Air Speed of sound depends on temperature v m = 331 s 331 m/s is the speed of sound at 0 C T is the absolute temperature T 273 K

18 Example: thunderstorm Suppose that you hear a clap of thunder 16.2 s after seeing the associated lightning stroke. The speed of sound waves in air is 343 m/s and the speed of light in air is 3.00 x 10 8 m/s. How far are you from the lightning stroke?

19 Example: Given: v light =343 m/s v sound =3x10 8 m/s t=16.2 s Since v, we ignore the time required for the light >> vsound lightning flash to reach the observer in comparison to the transit time for the sound. Find: Then, ( )( ) 3 d 343 m s 16.2 s = m = 5.56 km d=?

20 20 Standing Sound Waves Consider a pipe open at both ends: The ends of the pipe are open to the atmosphere. The open ends must be pressure nodes (and displacement antinodes).

21 Standing Sound Waves 21 Pipe open at both ends Example: Flute

22 Pipe Closed at One End 22 Example: Clarinet

23 The Human Ear 23

24 Intensity of Sound Waves The intensity of a wave is the rate at which the energy flows through a unit area, A, oriented perpendicular to the direction of travel of the wave ΔE I = = A Δt P A P is the power, the rate of energy transfer Units are W/m 2

25 Various Intensities of Sound Threshold of hearing Faintest sound most humans can hear About 1 x W/m 2 Threshold of pain Loudest sound most humans can tolerate About 1 W/m 2 The ear is a very sensitive detector of sound waves

26 Intensity Level of Sound Waves The sensation of loudness is logarithmic in the human hear β is the intensity level or the decibel level of the sound I β =10 log I o I o is the threshold of hearing Threshold of hearing is 0 db Threshold of pain is 120 db Jet airplanes are about 150 db

27 Example: rock concert The sound intensity at a rock concert is known to be about 1 W/m 2. How many decibels is that? and who is this guy?

28 Example: Given: I 0 =10-12 W/m 2 I 1 =10 0 W/m 2 1. Use a definition of intensity level in decibels: Find: 1. β=? β = 10 log 10 = 10 log I I 10 0 = = 10 log ( 10 ) = 120dB Note: same level of intensity level as pain threshold! Normal conversation s intensity level is about 50 db. The guy is Klaus Meine (Scorpions)

29 Doppler Effect A Doppler effect is experienced whenever there is relative motion between a source of waves and an observer. When the source and the observer are moving toward each other, the observer hears a higher frequency When the source and the observer are moving away from each other, the observer hears a lower frequency Although the Doppler Effect is commonly experienced with sound waves, it is a phenomena common to all waves

30 Doppler Effect, Observer Moving ƒ' = v + v ƒ v o

31 Doppler Effect, Observer Moving The apparent frequency, ƒ, depends on the actual frequency of the sound and the speeds ƒ' = v + v ƒ v o v o is positive if the observer is moving toward the source and negative if the observer is moving away from the source

32 Doppler Effect, Source Moving ƒ' = ƒ v v v s Use the v s when the source is moving toward the observer and +v s when the source is moving away from the observer

33 Doppler Effect, both moving Both the source and the observer could be moving v + v o ƒ' = ƒ v vs Use positive values of v o and v s if the motion is toward Frequency appears higher Use negative values of v o and v s if the motion is away Frequency appears lower

34 Example: taking a train An alert phys 2130 student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 442 Hz when the train is approaching him and 441 Hz when the train is receding from him. From this he can find the speed of the train. What value does he find?

35 Example: Given: frequencies: f 1 =442 Hz f 2 =441 Hz sound speed: v=345 m/s Find: v=? With the train approaching at speed, the observed frequency is 345 m s m s 442 Hz = f = f 345 m s vt 345 m s vt As the train recedes, the observed frequency is 345 m s Hz 345 m s = f = f 345 m s ( vt) 345 m s + vt Dividing equation (1) by (2) gives, m s + vt = m s v and solving for the speed of the train yields v t = m s t (1) (2)

36 36 Example: An ambulance traveling at 44 m/s approaches a car heading in the same direction at a speed of 28 m/s. The ambulance driver has a siren sounding at 550 Hz. At what frequency does the driver of the car hear the siren? This is a Doppler effect problem with both source and observer moving: f v 28 m s o 1 1 v 343 m s o f v s s 44 m s 1 1 v 343 m s = = (550 Hz) = 580 Hz

37 37 Echolocation Sound waves can be sent out from a transmitter of some sort; they will reflect off any objects they encounter and can be received back at their source. The time interval between emission and reception can be used to build up a picture of the scene.

38 38 Example: A boat is using sonar to detect the bottom of a freshwater lake. If the echo from a sonar signal is heard s after it is emitted, how deep is the lake? Assume the lake s temperature is uniform and at 25 C. The signal travels two times the depth of the lake so the one-way travel time is s. From table 12.1, the speed of sound in freshwater is 1493 m/s. depth = vδt = ( 1493 m/s)( s) = 403 m

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