Title 10. Sound Waves.pdf ✅

Sound Waves 201 10 Sound Waves QUICK LOOK For transverse wave propagation medium must be rigid. Air has no rigidity; therefore sound waves in air are longitudinal. In longitudinal waves pressure and density vary. Figure: 10.1 Speed of longitudinal wave E v d = where E = elasticity and d = density of medium In solids E v d = , in liquids K v d = and in gases P RT v d M γ γ = = Pressure has no effect on velocity of sound waves. Velocity of sound in a gas increases with increase of temperature i.e., v T ∝ , Velocity of sound decreases with increase of molecular weight 1 v M ∝ Velocity of transverse wave in a string 2 T T v m πr d = = , Where M m L = Mass per unit length and 2 M r Ld = π Note For small temperature difference t 0 0 546 t v v v t = + (for any gas), In air 0 0.61 m/s t v v t = + Equation of plane progressive wave sin( ) sin2 y A t kx t x A T ω π λ = −     = −      or sin( ) sin2 y A t kx t x A T ω π λ = +     = +      Particle velocity cos ( ) dy u A t kx dt = = − ω ω Maximum particle velocity, max u A = ⋅ ω Wave velocity v n = λ Slop dy u dx v = − Relation between path difference and phases difference 2 x λ φ π ∆ = × ∆ Excess pressure in a longitudinal wave dy u p E E dx v = − = Intensity of a wave, 2 2 2 I n A d v = ⋅ 2π Watt/m 2 . Echo: An echo can be cited as an example of reflection of sound from a distant object such as hill or cliff. It is basically a sound of short duration reflected back to the observer 0.1 sec or more after the production of original sound. If there is a sound reflector at a distance d form the source, the time interval between original sound and its echo at the site of source will be: d d d2 t v v v = + = Now as persistence of ear is (1/10) sec, echo of a sharp or momentary sound will be heard if 1 10 t > or 2 1 10 d v > i.e., 20 v d > If a person standing between two parallel hills fires a gun and hears the first echo after 1 t sec, the second echo after 2 t sec, and v is the velocity of sound, then the distance between the two hills is given by: ( 1 2 ) 1 2 1 2 2 2 2 vt vt v t t s s   + + = + =     A man standing in front of a mountain at a certain distance beats a drum at regular intervals. The drumming rate is gradually increased and he finds that the echo is not heard when the rate becomes n1 per minute. He then moves nearer to the mountain by x m and finds that the echo is again not Wave propagating along negative X-axis wave propagating along positive X-axis Transverse wave Compression Expansion Longitudinal wave Wavelength Wavelength
202 Quick Revision NCERT-PHYSICS heard when the drumming rate becomes n2 per minute. Then the distance between the mountain and the initial position is given by the equations 2 1 2 n n x d n   − =     or 2 1 2 1 1 2 n x xt d n n t t = = − − Where ( 1 2 ) 2 v t t x − = If a motor car approaching a cliff with a velocity u m/s sounds the horn and the echo is heard after t sec, then the distance between the cliff and the point where the horn is sounded is given by: 2 v u s t   + =     ( v = velocity of sound) The distance between the cliff and the point where the echo is heard is given by: 1 2 v u s t   − =     A road runs midway between two parallel rows of building. If a motorist moving with a speed u m/s sound the horn and hears the echo after t sec, then the distance between the two rows of building is 2 2 d v u t = − [ ] where v is the velocity of sound. A road runs parallel to a long line of smooth cliffs. If a motorist moving with speed u m/s sounds the horn and hears the echo after t sec, then the distance between the road and the cliffs is given by 2 2 [ ] 2 t d v u = − where v is the velocity of sound. Reflection: If incident wave is y A t kx = − sin ( ), ω then the equation of reflected wave is y rA t kx = + sin ( ) ω (from free boundary) = + + rA t kx sin ( ) ω π (from rigid boundary) where r reflection coefficient. Refraction: 1 2 sin sin i v r v = = constant Critical angle: ( ) 1 1 2 2 sin v C v v v = = < Condition of maxima, Path diff. ∆ = mλ Condition of minima, Path diff ( ) 2 1 2 n λ ∆ = − Intensity at a point where phase difference is δ 2 2 1 1 1 2 I a a a a = + + 2 cos δ 1 2 1 1 I I I I I = + + 2 ( ) cosδ Maximum intensity, 2 max 1 2 I a a = + ( ) Minimum intensity, 2 max 1 2 I a a = − ( ) Figure: 10.2 Beats: In beatstwo sources are coherent but of slightly different frequencies. Number of beats / sec 1 2 = − n n . Time interval between successive maximum (or successive minima) 1 2 1 T n n = − Equation of stationary wave is 2 2 2 sin cos x t y a T π π λ = (node x = 0) 2 2 2 sin cos x t a T π π λ = (antinode at x = 0) All particles between two consecutive nodes vibrates in same phase while the particles on opposite side of a node vibrate in opposite phase. Separation between two successive nodes (or antinode) = λ/2 At nodes displacement of particles is always zero so they are permanently at rest. But strain dy dx       at nodes is maximum, so pressure and hence energy is maximum at nodes. At antinodes the displacement is maximum and strain dy dx is zero; so pressure and hence energy is minimum at antinodes. Vibration of Strings, Rods and Air Columns Closed Organ Pipe The waves formed in a vibrating air column of a closed organ pipe are longitudinal stationary waves. In a closed organ pipe, the closed end is always a node while the open end is always an antinode. In the simplest mode of vibration, a node will be formed at the closed end and an antinode is formed at the open end. In the simplest node of vibration the length of the air column is equal to 1 4   λ     or 1 4 L   λ =     ⇒ 1λ = 4 . L Fundamental free frequency 4 v n l = f1 f2 fb Resulting condensation Resulting rarefaction Resulting condensation Constructive interference Constructive interference Constructive interference
Sound Waves 203 Figure: 10.3 Open Organ Pipe Fundamental frequency, 2 v n l = Ratio of frequencies of harmonics produced = 1 : 2 : 3 (all harmonics even and odd are produced) The maximum possible wavelength is equal to 2L and the other wavelengths are given by: (2 / ), L k k = 1, 2, 3... Figure: 10.4 Air Column: At first resonance, 1 4 L e λ + = Figure: 10.5 At second resonance, 2 3 4 L e λ + = ∴ 2 1 3 4 4 1 L L e λ λ λ − + = − = or λ = − 2(L L 2 1 ) Velocity of sound v n n L L = = − λ 2 ( 2 1 ) where n is the frequency of the tuning fork. End correction, 2 1 3 2 L L e − = End correction: e r D = = 0.6 0.3 , Where, D = diameter r = radius. Vibrations of Strings: (string clamped at both ends) Fundamental frequency 1 2 2 v T n l l m = = Figure: 10.6 Ratio of frequencies of harmonics produced = 1 : 2 : 3 : ..... If string vibrates in p-loops, pth harmonic is produced; frequency of pth harmonic 2 p p T n L m = Where p = 1, 2, 3, ... Figure: 10.7 Melde’s law p T = constant. Figure: 10.8 Harmonics and Overtones: The frequency of pth harmonic or (p – 1)th Table showing characteristic of different harmonics or overtones Node Standing Wave Tension (T) Electric vibrator Sonometer box Wires Load Fundamental n 1 f1 f2 = 2f1 2 Fn = nf1 3 Vibrating string analog A N A N A N N = node , A = antinode λ/4 5 4 λ 3 4 λ Air Water Third harmonic Fifth harmonic First harmonic = fundamental ←[motion o air molecules] 1 1 2 L = λ 1 2L f υ = L = λ2 2 1 2 L f f υ = = 3 3 2 L = λ 3 3 1 2 3 L f f υ = = (a) Pressure variation in the air node antinode L node Overtones (a) Tube open at one end Displacement of air node antinode L Overtones (a) Tube closed at one end Displacement of air Third harmonic Fifth harmonic First harmonic = fundamental 3 1 3 3 4 f f L υ = = 3 3 4 L = λ 5 5 4 L = λ 5 1 5 5 4 f f L υ = = 1 1 4 L = λ 1 4 f L υ = (b) Pressure variation in the air L L
204 Quick Revision NCERT-PHYSICS Table 10.1: Harmonic and Frequency Harmonic Mode Number of loops Number of A modes Number of modes Frequency Wave length First Fundame ntal 1 1 2 1 n n = (2L / 1) Second 1st Overtone 2 2 3 2 n n = (2L / 2) Third 2nd Vertone 3 3 3 3 n n = (2L / 3) Pth Harmonic (p –1)th vertone p p (p + 1) p n n = (2L / p) Kundt’s Tube: Kundt’s tube enables us to determine the velocity of sound in solids in the form of rods and also in gases. Figure: 10.9 If Lr be the length of rod, λr be the wavelength of sound waves produced in it and velocity of sound in the rod be vr , then 2 r Lr λ = or 2 λr r = L and 2 r r r v n nL = = λ If the mean distance between two nodes be La and wavelength be λa then 2 a La λ = or 2 λ a a = L Also, 2 [ a a a a v n nL v = = = λ velocity of sound in air] ∴ 2 2 r r r a a a v nL L v nL L = = or Velocity of sound in rod Velocity of sound in air = Length of the rod distance between twoconsecutive nodes = Comparison of Velocity of Sound in Different Gases: Let v1 and v2 are the velocities of sound in the two gases and the distance between two nodes are L1 and L2. Then, 1 1 1 2 2 2 2 2 v L n L v L n L = = Doppler’s Effect in Sound: When a source of sound or an observer or both are in motion relative to air, there is an apparent change in the frequency of sound as heard by the observe. This phenomenon is called Doppler’s effect. When a sounding source approaches an observer or a moving observer approaches a source of sound, the apparent frequency (or pitch increases.) When a sounding source recedes an observer or a moving observer recedes a sounding source, the apparent frequency (or pitch decreases.) Figure: 10.10 Case (i): Moving source approaches a stationary observer: The apparent wavelength λ ' is given by ' . ., ' s v v i e v λ λ λ λ − = × < The apparent frequency will be: ' . ., ' s v n n i e n n v v = × < + Note When source in motion crosses a stationary observer, the change in frequency 2 2 2 s s v v n n v v = − But if 2 , s s v n v v n v << ∆ = Case (ii): Moving source recedes from a stationary observer: The apparent wavelength λ ' is given by: ' s v v v λ λ + = × i.e., λ λ ' = The apparent frequency will be: ' s v n n v v = + i.e., n n ' < The fractional change in wavelength is: ' s v v λ λ λ − = The fractional change in frequency is : ' s s n n v n v v − = + Case (iii) Observer moving towards a stationary source: The apparent frequency is given by: ' , o v v n n v + = × i.e., n n ' > The wavelength observed by the observer will be: velocityof waverelative toobserver λ'= frequencyobserved byobserver ' ' o v v v n n λ λ + = = = It is worth noting here that when the source is stationary, the apparent wavelength λ ' λ 'is the same as λ The fractional change in frequency is : ' o n n v n n v − = × V Observer S Source λ λ Source Observer λ VS Stationary disk Wooden dowel Clamp Vibrating disk Copper rod Resonating chamber N N N N A A A loop loop