Chapter-16 Waves-I

Chapter-16 Waves-I Topics to be studied: Types of waves. Amplitude, phase, frequency, period, propagation speed of a wave Mechanical waves propagating along a stretched string. Wave equation Principle of superposition of waves Wave interference Standing waves, resonance

Ch 16-2,3 Waves: Mechanical Waves Wave Motion: Disturbance of particles of a medium 3 types Of Wave 1-Mechanical Waves: waves requires a medium to propagate, e.g. water & string waves 2-Electromagnetic waves : waves travel without a medium e.g. radio waves, etc 3-Matter waves: waves associated with atoms and subatomic particles Mechanical Waves: Two types 1-Transverse Waves- particle displacement  to wave velocity direction e.g. string waves 2-Longitudinal Wave: Particle displacement  to wave velocity direction e.g. Sound wave

Ch 16-4 Wavelength and Frequency Equation of a Wave: traveling towards right : y1(x,t)=ymsin(kx-t) traveling towards left: y2(x,t)=ymsin(kx+t); T is period of oscillations and related with angular frequency  by : =2/T Number of periods in 2 seconds  is wavelength and related with angular wave number  by: k=2/ k = Number of wavelengths in 2 meters

Checkpoint-Ch-16-1 The figure is composed of three snapshots, each of a wave traveling along a particular string. The phases for the waves are given by (a) 2x-4t (b) 4x-8t © 8x-16t. Which phase corresponds to which wave? Phase  = kx-t, where wave number k= 2/ and wave length  = 2 /k. The for (a)  = 2 /k =  (b)  = 2 /k = /2 (c)  = 2 /k = /4 Which phase corresponds to which wave? (a)=2 (b) = 3 (c) = 1

Ch16-5 Speed of a Traveling Wave Wave travels a distance of x in time t, then wave speed v=dx/dt : Since particle A retains its displacement, which means kx- t= constant Then v=+dx/dt= /k v =/T =f Positive v for kx- t Negative v for kx+ t

Ch16-6 Wave Speed on a stretched string Net Force on small length l is FR ; where FR=2( sin)=  (2) s=R’ ; 2= l /R; m =l; where  is linear mass density FR= m v2/R= (l/R) v2=(l/m )= / v= /

Ch16-6 Transverse velocity of particles on a stretched string Transverse velocity of string particles u=dy/dt where y(x,t)=ymsin(kx-t); u= -ymcos(kx-t); Transverse acceleration ay= du/dt =-2ymsin(kx-t) =-2y ; Maximum Transverse velocity um=-ym Maximum Transverse acceleration ay-m=- 2ym

Checkpoint-Ch-16-2 v=/; umax=-ym Here are the equations of three waves. (1) y(x,t)= 2 sin (4x-2t) (2) y(x,t)= sin (3x-4t) (3) y(x,t)= 2 sin (3x-3t) Rank the waves according to their (a) wave speed (b) maximum transverse speed, greatest first. v=/; umax=-ym (1) v=-2/4=-0.5 m/s umax =4 m/s 2) v=-4/3=-1.33 m/s 3) v=-3/3=-1.0 m/s umax =6 m/s

Ch16-6 Energy and Power of a Wave traveling along a String Kinetic energy K due to transverse velocity u and elastic potential energy U due to stretching of the string element At point a, element at rest (turning point), y=U=0 and K=0. At point b, element fully stretched y, U and K are max

Ch16-6 Energy and Power of a Wave traveling along a String Rate of Energy Transmission: Kinetic energy dK of a string segment with mass dm dK=dmu2/2=dx[-ymcos(kx-t)]2/2 dK=2ym2cos(kx-t)2 dx/2 dK/dt=v2ym2cos(kx-t)2/2 (dK/dt)Avg=[v2ym2 cos(kx-t)2/2]Avg [cos(kx-t)2]Avg=1/2 Then (dK/dt)Avg= v2ym2 /4 (dU/dt)Avg= (dK/dt)Avg Pavg= 2(dK/dt)Avg=v2ym2 /4

Ch16-9 Principle of Superposition for Waves Overlapping of two traveling waves y1(x,t) and y2(x,t) results in wave y’(x,t), the algebraic addition of the two waves: y’(x,t)= y1(x,t)+y2(x,t)

Ch16-10 Interference of Waves Interference : Resultant displacement of medium particle due to addition of two waves : Two traveling waves: y1 (x,t) = ymsin(kx- t) and y2 (x,t) = ymsin(kx-t+ ) interfere to give resultant wave y’(x,t)=ymsin(kx- t)+ymsin(kx- t+ ) sin + sin = 2sin{(+)/2}cos{(-)/2} Then y’(x,t)=[2ymcos(/2)] sin(kx- t+ /2) y’(x,t)=y’m sin (kx-wt+ /2) y’m= [2ymcos(/2)]

Ch16-12 Standing Waves Standing Waves: Due to interference of two identical traveling waves moving in opposite direction Fully constructive interference for time difference t= 0, T/2, T Fully destructive interference for time difference t= T/4, 3T/4

Ch16-12 Standing Waves y’(x,t)=ymsin(kx-t)+ymsin(kx+t) Two traveling waves moving in opposite direction : y1 (x,t) = ymsin(kx-t) and y2 (x,t) = ymsin(kx-t) interfere to give resultant standing wave y’(x,t)=ymsin(kx-t)+ymsin(kx+t) sin + sin = 2sin{(+)/2}cos{(-)/2} Then y’(x,t)= [2ym sin(kx)] cos(t) =y’m cos(t) where y’m= [2ym sin(kx)] and it depends upon x Nodes y’m=0 for kx=0,,..=n (n=0,1,2,..); x= n/k =n/2 Antinodes y’m=2ym for kx=/2,3/2..=(n+1/2); (n=0,1,2,..); then x= (n+1/2)/k = (n+1/2) /2

Ch16-13 Standing Waves and Resonances For certain frequencies, called resonance frequencies, traveling waves interferes and produce very large amplitude. The string will resonate if fixed end of the string are nodes: The first mode is 1/2=L ; 2nd mode is 2 (2/2) =L ; 3rd mode is 3 (3/2) =L Then nth mode n (n/2) =L (n=1,2,3,….) n =2L/n but fn=v/ n ; Then fn=nv/2L =n(v/2L)=nf1 n is called harmonic ; n=1 is called first harmonic and is called fundamental mode of oscillation and f1 is called fundamental frequency n=2 second harmonic etc A B

Suggested problems: Chapter 16   The quiz questions will be same or very similar to the following text-book problems. Refer to the course website for the latest version of this document. You are encouraged to seek the help of your instructor during his office hours. 5. A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.170 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.40 m; what is the wave speed? Answer: (a) 0.680 s; (b) 1.47 Hz; (c) 2.06 m/s 12. A The function y(x, t) = (15.0 cm) cos (πx – 15.0 πt), with x in meters and t in seconds, describes a wave on a taut string. What is the transverse speed for a point on the string at an instant when that point has the displacement y = +12.0 cm? Answer: 424 cm/s = 4.24 m/s 16. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 N. To what value must the tension be changed to raise the wave speed to 180 m/s? Answer: 135 N 32. What phase difference between two identical traveling waves, moving in the same direction along a stretched string, results in the combined wave having an amplitude 1.50 times that of the common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, and (c) wavelengths. Answer: (a) 82.8 º; (b) 1.45 rad; (c) 0.230 λ 49. A nylon guitar string has a linear density of 7.20 g/m and is under a tension of 150 N. The fixed supports are distance D = 90.0 cm apart. The string is oscillating in the standing wave pattern shown in Fig.16-38. Calculate the (a) speed, (b) wavelength, and (c) frequency of the traveling waves whose superposition gives this standing wave. Answer: (a) 144 m/s; (b) 60.0 cm; (c) 241 Hz 52. A rope, under a tension of 200 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m) sin (π x/2) sin(12π t), where x = 0 at one end of the rope, x is in meters, and t is in seconds.What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation? Answer: (a) 4.0 m; (b) 24 m/s; (c) 1.4 kg; (d) 0.11 s 76. A standing wave results from the sum of two transverse traveling waves given by y1 = 0.050 cos (πx – 4πt) and y2 = 0.050 cos (πx + 4πt), where x, y1, and y2 are in meters and t is in seconds. (a) What is the smallest positive value of x that corresponds to a node? Beginning at t = 0, what is the value of the (b) first, (c) second, and (d) third time the particle at x = 0 has zero velocity? Answer: (a) 0.50 m; (b) 0; (c) 0.25 s; (d) 0.50 s