Lesson 11 Physics 168 1
Oscillations and Waves 2
Simple harmonic motion If an object vibrates or oscillates back and forth over same path each cycle taking same amount of time motion is called periodic Mass and spring system is useful model for a periodic system 3
Simple harmonic motion (cont d) We assume that surface is frictionless There is a point where spring is neither stretched nor compressed Equilibrium position We measure displacement from that point (x = 0) Force exerted by spring depends on displacement F = kx 4
Simple harmonic motion (cont d) Minus sign on force indicates that it is a restoring force it is directed to restore mass to its equilibrium position k is spring constant Force is not constant so acceleration is not constant either 5
Simple harmonic motion (cont d) Displacement is measured from equilibrium point Amplitude is maximum displacement A cycle is a full to-and-from motion Period is time required to complete one cycle Frequency is number of cycles completed per second 6
Simple harmonic motion (cont d) Any vibrating system where restoring force is proportional to negative of displacement is in simple harmonic motion (SHM) and is often called a simple harmonic oscillator We know that potential energy of a spring is given by PE = 1 2 kx2 Total mechanical energy is then: E = 1 2 mv2 + 1 2 kx2 Total mechanical energy will be conserved as we are assuming system is frictionless 7
Energy in simple harmonic oscillator If mass is at limits of its motion energy is all potential If mass is at equilibrium point energy is all kinetic We know what potential energy is at turning points 8
Period and sinusoidal nature of SHM Figure shows how to get experimentally versus t for a mass on a spring A marking pen is attached to a mass on a spring and paper is pulled to left As paper moves with constant speed pen traces out displacement x x as a function of time General equation for such curve is x = A cos (!t + ) 2 f Phase constant 9
Sinusoidal Nature of SHM Consider an object on a spring on a frictionless surface Equilibrium F x = kx Using Newton s second law m d2 x dt 2 = kx General solution is x = A cos (!t + ) 10
Sinusoidal Nature of SHM (cont d) Velocity and acceleration can be calculated as function of time Displacement x v = v max sin!t v max = A (k/m) 1 2 Velocity v a = a max cos (2 t/t ) a max = k A/m Acceleration a 11
Spider Web A spider of mass 0.3 g waits in its web of negligible mass A slight movement causes the web to vibrate with a frequency of about 15 Hz a) Estimate value of spring stifness constant k for web (b) At what frequency would you expect web to vibrate if an insect of mass 0.1 g were trapped in addition to spider? 12
Frequency of SHM is given by = 1 r k 2 m ) k =(2 )2 m =2.7 N/m For m =4 10 4 kg we have = 1 r k 2 m = 13 Hz 13
Oscillating systems: object on a vertical spring X Fy = ky + mg Changing variables y 0 = y y 0 X Fy = k(y 0 + y 0 )+mg But ky 0 = mg! X F y = ky 0 From Newton s second law ky 0 = m d2 y dt 2 dy = dy 0 Solution is y 0 = A cos (!t + )! =(k/m) 1 2 d 2 y 0 dt 2 = k m y0 14
For small Oscillating systems: simple pendulum where arc length Repeatedly differentiating on both sides of d 2 s dt 2 Substituting and re-arranging gives gives Note that motion of pendulum does not depend on its mass! sin d 2 mg sin dt 2 g L d 2 General solution for small oscillation = m d2 s dt 2 s = L = L d2 dt 2 dt 2 = g L sin s 1 = 0 cos(!t + ) Where! 2 = g L and T = 2! =2 (L/g) 1 2 15
sin at small angles (Degrees) (Radians) sin % Difference 0 0 0 0 1 0.01745 0.01745 0.005 % 5 10 15 20 0.08727 0.08716 0.17453 0.17365 0.26180 0.25882 0.34907 0.34202 0.1% 0.5% 1.1% 2.0% 30 0.52360 0.50000 4.7% 16
Energy in a simple harmonic motion U = 1 2 kx2 U = 1 2 ka2 cos 2 (!t + ) K = 1 2 mv2 K = 1 2 m!2 A 2 sin 2 (!t + ) Using! 2 = k/m E = U + K = 1 2 ka2 [cos 2 (!t + )+sin 2 (!t + )] = 1 2 ka2 17
Wave Motion A mechanical wave is caused by a disturbance in a medium As wind passes over water's surface friction forces it to ripple Strength of wind, distance wind blows, and duration determine how big ripples will become Crest is highest point on wave Trough is lowest point on wave Wavelength is horizontal distance either between crests or troughs of two consecutive waves Wave height is vertical distance between wave's crest and next trough Wave period measures size of wave in time 18
Wave Motion (cont d) If you have ever watched ocean waves moving toward shore before they break you may have wondered if waves were carrying water from far out at sea into beach Water moves with a recognizable velocity but each molecule of water itself This is clearly demonstrated by observing a bottle on a pond as waves move by Bottle is not carried forward by waves, but simply oscillate about an Water droplet move in a vertical circle as wave passes Droplet moves forward with wave's crest and backward with trough These vertical circles are more obvious at surface merely oscillates about an equilibrium point equilibrium point because this is motion of water itself As depth increases their effects slowly decrease until completely disappearing about half a wavelength below surface 19
Looking a little more closely at how a wave its form and how it comes to travel: A single wave bump or pulse can be formed on a rope by a quick up-and-down motion of hand Figure shows a pulse on a string at time Shape of string at this instant can be represented by some function String is described in this frame by At some later time pulse is farther down string for all times x-coordinates of two reference frames are related by Shape of string in original reference frame is f(x 0 )=f(x Pulse In a new coordinate system with origin that moves to right with same speed as pulse f(x 0 ) Same line of reasoning for a pulse moving to left leads to pulse is stationary x 0 = x wave moving in t =0 y = f(x) direction 20 vt) +x vt O 0 y = f(x + vt)
Looking a little more closely at how a wave its form and how it comes to travel: Periodic wave A continuous or periodic wave has as its source a disturbance that is continuous and oscillating disturbance in this case is change in shape of string Its propagation arises from interaction of each string segment with adjacent segments Segments of string move in direction perpendicular to string as pulses that propagate back and forth along string 21
Types of waves Waves in which motion of medium (molecules of water, particles on string) is perpendicular to direction of propagation are called transverse waves Waves in which motion of medium is along (parallel to) direction of propagation of disturbance are called longitudinal waves (Sound waves are examples of longitudinal waves) 22
Speed of waves Speed of waves relative to medium depends on elastic and inertial properties of medium but is independent of motion of source of waves String tension For a pulse on a rope v = FT µ 1 2 Linear mass density Bulk modulus For sounds waves v = B 1 2 Volume mass density 23
Graphic representations of a sound wave (A) Air at equilibrium in absence of a sound wave (B) Compressions and rarefactions that constitute a sound wave (C) Transverse representation of wave showing amplitude (A) and wavelength (λ) 24
Speed of sound in air For sound waves in a gas bulk modulus is proportional to pressure which in turn is proportional to density and to absolute temperature of gas Ratio B/ is independent of density and is merely proportional to absolute temperature Dimensionless constant For diatomic molecules such as O 2 depends on kind of gas Because and comprise of atmosphere for air For gases composed of monoatomic molecules such as Molar mass from air is Speed of sound at v =( RT/M) 1 2 25 O 2 N 2 N 2 98% =7/5= 7/5 H e =5/3 T [K] = T [ C] + 273 R =8.3145 J/(mol K) M = 29 10 3 kg/mol 20 C is about 343 m/s and
Explosion of a depth charge beneath surface of a body of water is recorded by an helicopter hovering above water's surface as shown in figure Along which path (A, B, or, C) will sound wave take least time to reach helicopter? 26
Explosion of a depth charge beneath surface of a body of water is recorded by an helicopter hovering above water's surface as shown in figure Along which path (A, B, or, C) will sound wave take least time to reach helicopter? Speed of sound in water is greater than speed of sound in air path C fluis are almost incompressible implies disturbance propagates very quickly 27
Wave intensity Wave transports energy from one place to another As waves travel through a medium energy is transferred as vibrational energy from a particle to particle in medium If a point source emits waves uniformly in all directions then energy at a distance r from source is distributed uniformly on a spherical surface of radius r and area A =4 r 2 Average power per unit area that is incident perpendicular to direction of propagation is called intensity I = hp i A 28
Doppler Effect You may have noticed that you hear higher pitch of whistle on a speeding train dropped abruptly as it passes you This phenomenon is known as Doppler effect 29
Doppler Effect (cont d) Consider whistle of a train at rest which is emitting sound of a particular frequency in all directions as shown in figure Sound waves are moving at speed of sound in air which is independent of velocity of source or observer If our source is moving then whistle emits sounds at same frequency as it does at rest Sound wavefronts it emits forward are closer together than when train is at rest This is because train as it moves is chasing previously emitted wavefronts and emits each crest closer to previous one Thus an observer in front of train will detect more wave crests passing per second so frequency heard is higher Wavefronts emitted behind train are farther apart than when train is at rest because train is speeding away from them Fewer wave crest per second pass by an observer behind moving train and perceived pitch is lower 30
Doppler Effect (cont d) In following discussion all motions are relative to medium Consider a source moving with speed u s and a stationary receiver Source has frequency Received frequency (number of crests passing receiver per unit time) is related to wavelength (distance between successive crests) and wave speed v by f s f r = v 31
Time between these two events is Doppler Effect (cont d) t 1 and next wave crest leaves source at time t 2 T s = t 2 t 1 A wave crest leaves source at time and during this time source and crest leaving source at time t 1 u s T s vt s travel distances and respectively At time t 2 distance between source and crest leaving at time t 1 equals wavelength 32
Doppler Effect (cont d) If u s <v Behind source We can express both = b =(v + u s ) T s In front of source Substituting for our expression for f r = v = = f =(v u s ) T s b 33 and v v ± u s f s f as =(v ± u s )T s = v ± u s f s and rearranging (stationary receiver)
Doppler Effect (cont d) When receiver moves with respect to medium received frequency is different because receiver moves past more or fewer wave crests in a given time Let T r denote time between arrivals of successive crests for a receiver moving with speed u r During time between arrivals of two successive crests each crest will have traveled a distance vt r and during same time receiver will have traveled a distance u r T r If receiver moves in direction opposite to wave during a time distance a wave moves + distance receiver moves equals wavelength vt r + u r T r =! T r = /(v + u r ) If receiver moves in same direction as wave vt r u r T r =! T r = /(v u r ) 34 T
Doppler Effect (cont d) Because f r = /T Substituting for we have f = 1 T r = v ± u r r f r = v ± u r v ± u s f s In a reference frame in which medium is moving (e.g. reference frame of ground if air is medium and there is a wind blowing) Wave speed v is replaced by v 0 = v ± u w where u w is velocity of wind relative to ground 35
Batman has sent a signal to batcave calling for his batfriends to cover his escape Answering signal, a bat which is nearby starts flying at 5 m/s As it flies, bat emits an ultrasonic sound wave with frequency 30 khz towards tall wall of building What frequency does bat hear in reflected wave? 36
The wall is treated as stationary observer for calculation frequency it receives Bat is flying towards wall so f 0 wall = f bat 1 1 v bat /v sound Wall is treated as stationary source emitting frequency f 0 wall and bat as moving observer flying towards wall f 00 bat = f 0 wall 1+ v bat v sound = f bat 1 1 v bat /v sound v sound + v bat = f bat =3.09 10 4 Hz v sound v bat 1+ v bat v sound 37
Shock Waves During our derivations of Doppler-shift expressions we assumed that speed u of source was less than wave speed v If source moves with speed greater than wave speed there will be no waves in from of source Instead waves pile-up behind source to form a shock wave In case of sound waves this shock wave is heard as a sonic boom when it arrives at receiver 38
Shock waves produced by a bullet traversing a helium balloon 39
Mach Number P 1 P 1 Figure shows a source originally at point After some time t wave emitted from point Source has traveled a distance moving to right with velocity u has traveled a distance ut and will be at point Line from this new position of source to wavefront emitted when source was at P 1 vt P 2 makes an angle with path of source known as Mach angle + sin = vt ut = v u Shock wave is confined to a cone that narrows as u v u increases Ratio of source speed to wave speed is called Mach number Mach number = u v 40
Sonic Boom Shock waves from a supersonic plane Luis Anchordoqui 41
A supersonic plane flying due east at an altitude of 15 km passes directly over point P Sonic boom is heard at point P when plane is 22 km east of point P What is speed of plane? 42
A supersonic plane flying due east at an altitude of 15 km passes directly over point P Sonic boom is heard at point P when plane is 22 km east of point P What is speed of plane? Velocity = 610 m/s 43
Doppler Effect (summary) Stationary Sound Source Doppler Shift Source moving with v source <v sound Breaking Sound Barrier Sonic Boom Source moving with v source >v sound 44