Chapter B Wave Motion Learning Objectives Introduction of the wave, wave velocity and particle velocity, types of waves and their applications, speed of the wave in the stretched string, energy, power and intensity of plane progressive wave, standing wave and resonance, sonometer. : (4 Hours) UCM ole El-} a, Time period (T): T = “5. where f is frequency. de Group velocity (v,): v, = ac. where o is angular velocity and k is wave number. 2n 2 z Phase difference (Ad): Ao = aE he = she X,— xX) = x At, where At is the time interval. Wave equation: v = x f, where v is wave velocity, A is wavelength and fis frequency. . di Particle velocity: u = a =ova-y where @ is the angular frequency, a is amplitude and y is displacement. Velocity of stretched string: v = aff. where T is tension and pt is mass per unit length of the string. Progressive wave equation: y=asin wt (from origin) = asin (wt — 6) = asin (wt - kx) (from distance x from origin) th where a is amplitude and t is time; and © = 2nf with the frequency f of the vibration. = Intensity of sound: I = 2n'vp a? a where f is frequency, v is velocity, a is the explode a wave and p density of the medium.of sound: P = 2n” py fa?S =S pv? a? where S is the area of the vibrating medium and Per unit length of the string. mary wave equation: y = A sin wt where A = ary wave. Theoretical Question EVI -te What is a wave? 2. What is the Progressive equation. % . Progressive wave: A transverse wave (or compression and y =asin wt «(i where a is amplitude and t is time; and © = 2rf with the frequency of vibration f. Now, at a distance x from origin a O, the phase difference 6 is =2E x aks 2a Fig.: Progressive wave where k = is the wave number. The displacement of a particle at Ois given by y = asin (wt - 9) y =asin (wt-kx) _... (ii) This eq® (ii) is the mathematical form of th equation moving in the positive x-direction, Point P at distance x from origi ‘ ve © progressive wa¥Wave Motion () Law of mass: The frequency of transverse vibration of a stretched string is inversely proportional to the square root of the mass per unit length p of the stretched string when length Jand tension T are kept constant. So, 1 fe—=, when /and T are kept constant. Vi By combining these relations, we have ieee a ere tex} aft 120) where k is the proportionality constant, whose value is found mh tobe>. Hence, f = 4 aft «+ (ii) Numerical Problems . (20 | 1. - Asimple harmonic wave is represented by y = 10 sin (7 + ¢) The time period is 30 s. At time t = 0, the displacement is 5 cm. (a) phase angle at t = 7.5 s and (b) . the phase difference between two positions at a time interval of 6s. %= — Given, the given equation of wave is y= rosin (Et + ) Time period (T) = 30s Displacement at t = 0 (y) = 5cm Phase angle at time 7.5 s (') =? Phase difference at time interval 6 s (Ag) = ? Att = 0, we have y=asing $= sin’ () = sin’ @) = 30°= grad (a) Att = 7.5 s, the phase angle is 2: = rad40_A Reference Book on. lied Physics. Att =6s, the phase difference is 2 an ,, 28, g =Srad Ag =p At = 30 *® 5 (b) . 20 Hence, the value of phase difference at t = 7.5 s is 3 rad and the hay iti ime i _ 2m difference between two positions at the time interval of 6 s is Fad 2. The equation of a transverse wave travelling along a very lng stringis y = 6.0 sin (0.020 mx + 4.0 mt) t where x and y are expressed in centimetres and t is in sec. Determine: (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of th: wave and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x = 35 a when t = 0.26 s? : » Given, the displacement of the given wave is y = 6 sin (0.02.nx + 4nt) . (a) amplitude (y,,) =? (b) wavelength (A) =? (c) frequency (f) = ? (d) speed (v) =? () direction of wave =? — (f) maximum speed (Vn (g) displacement at x = 3.5 cm and t = 0.26 s (y) =? On comparing with the general equation of wave is Y = Yu Sin (kx + wt) 6 GH We get, y,, = 6 cm, k = 0.02 mrad/cm and w = 4x rad/s (a) Amplitude (y,) = 6 cm (b) Wavelength (4) = = _—2 0.02 7 = 100 cm () Fi ap = ae Tequency (f) = on = pp Hz On (@) Speed (v) =f2=2 x 100 = 200 cm/s wat (e) ba wave function being (kx + wt), the give? 0 eure along the negative x-direction. * 8) Form SPO Wand) = 0 Yq = 4 x 6 = 75 emis or new displacement atx = 3.5cmandt = 0.26 8 is Y&Q=y gi 4 oe Hie RD xX, t= . ‘06 = 6 sin (0.02 n x 3.5) + (4m x 0-20)! y: & 0 =6sin 1994" Y & 0) =-26m or,Wave Motion 44 Hence, the values of amplitude is 6 cm, the wavelength is 100 cm, frequency is 2 Hz, the speed is 200 cm, the wave propagates along the negative x-axis, the maximum speed is 75 cm/s and the new displacement of the wave at x = 3.5 cm and t = 0.26 s is -2 cm. Calculate the frequency and maximum particle velocity due to the wave represented by y (x, t) = 0.03 sin (607 t — 0.03 x). The values of x and y are measured in centimetres. Given, y = 0.03 sin (607 t — 0.03 x) iva Gi) Frequency (f) = ? Particle velocity (u) = ? Comparing with, y = asin (at - kx) : ove {UJ We get, a = 0.03 cm, @ = 60 m and k = 0.03 cm’ fo emir 2 OOme Using, f = on = 2n 7 30 Hz For maximum particle velocity, we have u = @a = 60 7 X 0.03 = 5.65 cm/s Hence, the value of frequency is 30 Hz and the maximum particle velocity due to the wave is 5.65 cm/s. Calculate the wavelength, frequency, speed of the wave and maximum particle velocity in the wave represented by y = 20 sin x (2t - 0.05 x) where the values of x and y are in centimeters. - [PoU 2013] Given, y = 20 sin m (2t — 0.05 x) = 20 sin (2nt- 0.05 x) Wavelength (A) =? Frequency (f) = ? Speed of the wave (v) =? Maximum particle velocity (Unad Coniparing with the progressive equation, y =x, sin (wt - kx) We get, x, = 20cm =0.2m, = 2nrad/s and k = 0,05ncm™ = 52 m™* =1 Using, f= 2 = 22 = 1 He 20 And, 4 = = 28 = 0.4m So, v=f£xi=1x04=04m/s And, Ups, = Xp 0 = 0.2 X 2m = 1.26 m/s ie Hence, the value of the wavelength is 0.4 m, the frequency 1s ‘i ia the speed of the wave is 0.4 m/s and the maximum particle velocity is 1.26 m/s,42_AReterence Book on Applied Physics 5. What is the amplitude, wavelength and velocity of the wan represented by y = 5 sin (Gxt + 4x), where distance and time oy, measured in S.I. units? [Pou 2o1y 3 Given, y = 5 sin (6nt + 4x) Amplitude (x,) =? Wavelength (A) = ? Velocity of wave (v) = ? Comparing with the progressive equation, y = x,, sin (wt - kx) We get,x,=5m, @=6nrad/s and k=4m" ‘igi po ee Me Using, f = 37 = pq =.3 Hz -2n_ 20 And, A= [== 157m So, v=f£xA=3 x 1.57 = 4.71 m/s Hence, the value of the amplitude is 5 m, the wavelength is 1.57 1 and the velocity of the wave is 4.71 m/s. 6. What will be the ratio of intensities of two waves y, = 5 sin (0.3 t- 20 x) and y, = 4 sin (0.4 t - 25 x) where the value of x and y are in cm and t is in second? = Given, y, = 5 sin (0.3 t- 20 x) y, = 4 sin (0.4 t-25 x) Ratio of intensities of two waves (I, Comparing with the progressive equation, y =asin (wt-ko) For the first wave, we have a, = 5m = 0.05 m, @, = 0.3 rad/s and k, = 20 cm* = 2000 m™ Os. 1.5 x 10 m/s For the second wave, we have a, = 4cm = 0.04 m, o, = 0.4 rad/s and k, = 25 cm™ = 2500 m* m2 04 8 4 onjs v, =, = 2500 7.16% 10 m/s Since the intensity of the wave is 1 l=7pvo'ape es 2 eye Motion” 5A3 _ pyofa/2_vofae _ 1.5 x 10" x 0.3? x 0,05? So, 1, = pv, @,7a,/2 ~ V, 7a," 1.6 X 10° X 0.4" x 0.047 I Te = 0.83 Hence, the ratio of the intensities of the given two waves is 0.83. 7. Find out the frequency of vibration of air particles in a plane progressive wave of amplitude 2.5 x 10° cm and intensity 10° W/m’, the speed of the sound in air as 332 m/s and density of air is 1.29 kg/m*. Given, amplitude (a) = 2.5 x 10% cm = 2.5 x 10m Intensity (I) = 10° W/m? Speed of the sound in air (v) = 332 m/s Density of air (p) = 1.29 kg/m® Frequency of vibration of air particle (f) = ? Since the intensity of the sound is 4 i l= Zpvo’ a’ = Zev (am fy’ a? = 2npvf? a” or, f= L__., | oe » F=\Vonpy a \j 2m x 1.29 x 332 x (2.5 x 10°)" f= 771 Hz Hence, the value of the frequency of vibration of air particles is 771 Hz. 8. A progressive and stationary, simple harmonic wave having a frequency of 250 Hz and each having the same velocity of 30 m/s. (a) Determine the phase difference between two vibrating points in a progressive wave at a distance of 10 cm. (b) Wave equation of progressive wave if the amplitude is 0.03 m. (c) Distance between nodes in a stationary wave. [PoU 2015] % Given, frequency (f) = 250 Hz . Velocity (v) = 30 m/s (a) Path difference (Ax) = 10 cm = 0.1m Phase difference (Ad) = ? (b) Amplitude (a) = 0.03 m Progressive wave equation (y)=? (c) Distance between nodes (2/2) = ? v_ 30 Here, 4 =F = 355 = 0.12 m (a) For progressive wave, we have : 20 2: é A = FE x Ax = G49 X 0-1 = 5.2444 ___A Reference Book on Applied Physics 10. (b) The progressive wave equation is y = asin (wt - kx) = a sin (ant t-28x) or, y= 0.03 sin (on X 250 x 1-28 x) y = 0.03 sin (5007 t - 52.4 x) (c) For the distance between two nodes, we have a = “2 = 0.06 m Hence, the value of phase difference is 5.24°, the progressive wavy equation is y = 0.03 sin (500n t — 52.4 x) and the distance betwee two nodes is 0.06 m. A wave of frequency 512 Hz has a phase velocity of 340 mi (a) How far are two points 30° out of phase? (b) Calculate the phase difference between two displacements at certain points at time of 5 ms apart. F Given, frequency (f) = 512 Hz Phase velocity (v) = 340 m/s (a) Phase difference (Ad) = 30° Path difference (Ax) = ? (b) Time period (AT) = 5 ms = 5 x 10° s Phase difference (Ao) = ? 512 © yee 3 Here, A= = 349 = 1,51 mand T = F = 579 = 1.95 x 10° s (a) For path difference, we have 2n Ao = es Ax a dibl Oh08 ea Ax = 55 x AQ =D X 30° = 7 yaggs = 0.126 m (b) For phase difference, we have 20 TOE x 103 * 5 X 107 = 161 04 0 gut Hence, the value of the distance between two points 30 a phase is 0.126 m and the phase difference between two poln time 5 ms is 16,1 rad. A metal wire with a mass of 5 gm and length of 100 cm is st°'y) | with a tension of 50 N. A wave with a frequency of 60 Ha amplitude of 2 mm travels along the wire. (a) Fi a fd average power carried by the wave. (b) What happe"> power of the wave if its amplitude is halved? 2 Ag = 3p x AT = ssWave Motion _45 ee Given, mass of wire (m) = 5 gm = 5 x 10° kg Length of wire () = 100cm =1m Tension (T) = 50N Frequency (f) = 60 Hz Amplitude (a) = 2mm = 2 x 10°m Average power (P) = ? New power for halved amplitude (P’) = 7 For mass per unit length of the wire, we have m_5x10° BET 7 7 = 5X 10% kg/m @ = 2nf = 2n x 60 = 120m rad/s T [ And, = [2 = Soe 100 m/s For average power, we have 1 1 = hv wa = 2% 5 x 10° x 100 x (120n)* x (2 x 10°)? P=0.142 W For halved amplitude, we have 2-82 2107 a=7=— 7. 7 10°m So, P= dav @ (a? = =+ 2% 5 x 10° x 100 x (120n)* x (10°)? P'= 3.55 x 107 W Hence, the average power carried by the wave is 0.142 W and the new Power carried by the wave as the amplitude is halved is 3.55 x 10° W. - Two progressive waves of equal amplitude and frequency travelling in opposite directions superimpose each other to form a standing wave of equation y = A cos kx sin wt, where A = 1 mm, k= 1.57 cm" and w = 78.5 s*. Find: (a) Velocity of a progressive wave. (b) Node closed to the origin, x > 0. (©) Antinode closed to the origin, x > 0. (4) Amplitude of resultant wave, when x = 2.33 cm. [PoU 2017 Spring] Given, y = A cos kx sin wt Amplitude (A) = 1mm = 10° m Wave vector (k) =1.57 cm? = 157 m™ © = 78,5 54 oat48_AReference Book on Applied Physics 12. (a) Velocity of wave (v) = (b) Distance between node closed to the origin (x) =? (c) Distance between anti-node closed to the origin (x’) =? (d) Amplitude of wave for x = 2.33 cm (Ag) = ? (a) For the velocity of the progressive wave, we have @ _ 785 er Kk 457, (b) As the amplitude of the node is zero, we can write = 0.5 m/s 1 cos kx = 0 = cos) x or, kk=> BE as kext (c) As the anti-node amplitude is maximum; we can write cos kx = 1 = cosm or, kk=" (d) For the amplitude of vibration, we have kx = 157 x 2.33 x 107 rad = 3.6 rad = 3.66 x ie kx = 210° Ag = Acos kx = 10° x cos 210° = -8.66 x 107m A source of sound has a frequency of 256 Hz and an ampli of 0.50 cm. Calculate the energy flow across a square meter sec. (Given, velocity of sound in air = 330 m/s and the density air = 1.29 kg/m’) [PoU 2021! Given, frequency (f) = 256 Hz Amplitude (a) = 0.5 cm = 0.5 x 107m Velocity (v) = 330 m/s Density (p) = 1.29 kg/m? Intensity or energy flow across a cm’/s (I) = ? We have, I = — sie =x4- ar pf atv or, I = 2m? x 1.29 x (256)* x (0.5 x 107)? x 330 I= 1.38 x 10* J/m*-s j Hence, the value of energy flow across a square cm per 5° 1.38 x 10* J/m’ss.Wave Motion 47 ee Se a, ae aa ee 43, A wave of frequency 500 Hz has a phase velocity of 200 m/s. 14. 15. Calculate: (a) How far apart are two points 30° out of phase? (b) What is the phase difference between two displacements at a point in time 10° s apart? Given, frequency (f) = 500 Hz Wave velocity (v) = 200 m/s Time taken (t) = 10° s Phase difference (6) = 30° = § Path difference (x) = ? New phase difference (¢') = ? (a) For path difference far apart two points 30° out of phase, we have, wpe Ber Sea al X= n° = Inf? = 2m x 500 * 6 > 233m (b) For phase difference between two displacements at a point at time 10° s apart, we have = ot = 2nft = 2x x 500 x 10° = mradian = 180° Hence, the values of the path difference of the given wave for 30° out of phase is 3.33 cm and the phase difference at time 10° s apart is 180°. A wave of frequency 500 Hz has a phase velocity of 350 m/s. How far are two points with a phase difference of 60°? Given, frequency (f) = 500 Hz Phase velocity (v) = 350 m/s x Phase difference (A$) = =3 Path difference (Ax) = ? Using, 4 = ¥ = 38 - 0.7m So, do = 22 x ax dx = AMG 07 293 _ gaz m Hence, the path difference between the two points is 0.117 m. A sinusoidal wave of frequency 500 Hz has a speed of 350 m/s. (a) How far apart are two points that differ by 3 rad? (b) What is the Phase difference between two displacements at a certain point at times 1.00 ms part?48 A Reference Book on Applied Physic w= Given, frequency (f) = 500 Hz Speed (v) = 350 my, Phase difference (6) = 5rad Path difference (x)=? New phase difference for t = 1 ms (9) = ? ¥ 350" 07 For wavelength, we have 4=F=500 7 0-7m 1 a1 For the time period, wehave T= F=5p9 = 2 x 10°s (a) For path difference, we have Xu ( 2n ) == o> ve«F, foraconstant p. % KF So, vy F, cA 180)? EB, =F, (3) = 120 C= = 134.5N Hence, the value of tension for the given wave is 134.5 N. A rod vibrating at 12 Hz generates harmonic waves with an amplitude of 1.5 mm in a string of linear density 2 gm/m. If the tension in the string is 15 N, what is the average power supplied by the source? Given, frequency (f) = 12 Hz Amplitude (a) = 1.5 mm = 1.5 x 10°m Linear density (u) = 2 gm/m = 2 x 10° kg/m Tension (T) = 15N Average power supplied (P,.) = ? Using, Pay =fuvod =3xu xa [Ex (anf)? x a? or, P,,= 20 fa’ (TI yu)” P,, = 20 Xx 12? x (1.5 x 10°)’ x (15 x 24 P,, = 1.11 x 10° W Hence, the average power supplied by the given source of sound is 1.11 x 10° W. Astring has a linear density of 525 gm/m and a tension of 45 N. When a sinusoidal wave of frequency 120 Hz and amplitude 8.5 mm is sent along the string, at what average rate does the wave transport energy? [PoU 2014 Spring] Given, linear density (4) = 525 gm/m = Tension (T) = 45N Frequency (f) = 120 Hz Amplitude (1) = 8.5 mm = 8. Average power (P,,) = ? 4 T 45, U mi [45_ = sing, V aft =\fo525 = 926 m/s So, the average power is P,, = 210 pv f7 P,, = 2n? x 0.525 x 9.26 x (120)° Hence, the average rate of energy transp the wave is 99.8 W. ve or, 0°) 0.525 kg/m 5x 10°m x (8.5 X 10°)’ = 99.8 W ort i.e. average power of50_A Reference Book on Applied Physies i itted by a transverse wave of 20. What power is transm i ; 4 kHz and amplitude 1 cm propagating through a vel Ci mass density 1 gm/m and under a tension of 20 N? ws Given, frequency (f) = 1 kHz = 10° Hz Amplitude (r) = 1cm = 0.01 m Linear mass density (1) = 1 gm/m = 10° kg/m Tension (T) = 20N Power (P) =? a 20 Here, v= ft =a] 5 141.4 m/s So, p=d7 ofp =3 F nf) x pv P= 2n' rf pv = 2m? x (0.01)*x(10°)?X10°X 141.4 = 279.1 Hence, the power transmitted by wave through the string is 279.1 W. 24. The elastic limit of steel forming a wire is equal to 2.70 x 10° Pe What is the maximum speed at which transverse wave pulses cat propagate along this wire without exceeding this stress? (Densil of steel = 7.89 x 10° kg/m’) ~ ~% — Given, elastic limit of steel (P) = 2.7 x 10° Pa Density of steel (p) = 7.89 x 10° kg/m* Speed (v) =? oye |Pie: [2z x 10° Using, v = fe =\J7.89 x 10° = 184.99 m/s Hence, the value of the maximum speed of the given wave propagating through the given steel wire is 184.99 m/s- 22. A simple harmonic transverse wave is propagating along ® sirist towards the left direction as shown in the figure. ae strowea Plot of displacement as a function of pes al time t = 0. The ati es ns densill age: Gulia ties mone S38 cod tint We wave speed, (d) the period, (e) the maxim icle speed string and (9 write an equation forte ance wae 018Wave Motion _51 w= Given, string tension (T) = 3.6 N Linear density (1) = 25 g/m = 25 x 10° kg/m (a) Amplitude (A) =? (b) Wavelength (A) =? (c) Wave speed (v) = ? (a) Time period (T) =? (e) Maximum particle speed (v,..,) = ? () Equation of the travelling wave (y) = ? {a) The amplitude of the wave is A=5cm=005m (b) The wavelength is A4=40cm=04m (c) The wave speed is a [ 3.6 v=nft- 25 x 103 = 12 m/s (d) The frequency is The time period is 1 1 Ta ¢ = 305 3.33 X 107s (ce) The maximum particle speed is Vmax = A@ = A’X 2nf = 0.05 x 2n X 30 = 9.4 m/s () The equation of the travelling wave is on 2 y= Asin ot + kx) = Asin (2-72) 3 20 2n Pia S es sin (335 x 107 1-24x) y = 0.05 sin (602 + 52 x) A piano wire with a mass of 5 gm and length of 90 cm is stretched with a tension of 25 N. A wave with a frequency of 100 Hz and an amplitude of 1.6 mm travels along the wire. Calculate the average Wnrer carciad by the wave and what happens if the power of the ee mplitude is halved. [PoU 2020 Fall] ven, mass of wire (m) = 5 gm = 5 x 10°kg Length of wire () = 90 cm = 0.9m Tension on wire (T) = 25 N Frequency (f) = 100 Hz Amplitude (a) = 1.6 mm = 1.6 X 10°m Average power (P.) =? New power for a/2 (P,.) =? ON err ices52 A Reference Book on lied Physics es Here, w= 7 = axe = 5:56 x 10° kg/m The speed of the wave in the wire is mies ie tea ve fE- 556 x 107 = 97-1 m/s The average power is 1 Ps =dpve? @ = qava’ (2m f?) = 2n? pv a? f? or, P,, = 2m’ X 5.56 X 10° x 67.1 X (1.6 x 10°)? x 100? wv = 0.189 W If the wave amplitude is halved, the power of the wave will be reduced by one-fourth. So, P, 0.189 P,' == z 2 4 4 — 471 10° W Hence, the value of the average Power carried by the wave is 0.189 W. 24, Calculate the speed of the transverse wave in a rope of length 20 cm having 60 gm of wire under a tension of 500 N. [PoU 2014 Fall] % Given, Length of rope ()) = 20cm = 0.2 m Mass of rope (m) = 60 gm = 0.06 kg Tension (T) = 500 N Speed of wave (v) = ? Hore, » = 77 = 9:98 _ 0.3 kg/m Now, the speed of the wave in the rope is iE 500 van fEay j200 40.8 m/s 25. A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 5: What are (a) the period and frequency? (b) the wavelength is 140 1: What is the wave speed? [PoU 2017] 3 Given, time for maximum displacement to zero (T/4) = 0.17 s Wavelength (A) = 1.4m Time period (T) = ? Frequency (f) = Wave speed (v) = ? The time period of the wave is T=4x (T/4)=4% 0.17 = 0.685Wave Motion 53 26. The frequency of the wave is oe f= T= O68 = 1:47 Hz The wave speed is v=fxA=1.47 X 1.4 = 2.06 m/s Hence, the value of the time period is 0.68 s, the frequency is 1.47 Hz and the wave speed is 2.06 m/s. Here are the three equations of the wave: y (x, t) = 2 sin (4x - 2t) y (x, t) = 2 sin (3x - 4t) and y (x, t) = 2 sin (3x - 3t) Rank the waves according to their (a) wave speed and (b) maximum transverse speed with the greatest first, [PoU 2021 Fall] Given, the three equations of the wave are y (x, t) = 2 sin (4x - 2t) y (x, t) = 2 sin (3x — 4t) and y (x, t) = 2 sin (3x — 3t) Comparing with the standard wave equation, y = asin (wt — kx) (a) For the first wave, we get @ = 4rad/s andk = 2 m"* The wave speed is v= e = $ =2m/s The maximum transverse speed is (Vmods = aM = 2X 4=8m/s (b) For the second wave, we get : = 3 rad/s and k = 4m™ The wave speed is The maximum transverse speed is (Vnad; = aw = 2X 3 =6 m/s (©) For the third wave, we get © = 3 rad/s and k = 3m™ The wave speed is 3 3 =1m/s g YW=RF54__A Reference Book on Applied Physics 27. The maximum transverse speed is (Wad: = a® = 2X 3 =6 m/s Hence, the rank of the three waves according to their wave speed is first, third and second whereas according to their maximum transverse speed is first is maximum while the remaining two are the same. A wave is propagating on a long stretched string along its length taken as the positive x-axis. The wave equation is given as y=y,e"™-", where y, = 4mm, T = 1.0s and A = 4m, Find: (a) the velocity of the wave; (b) _ the function f (t) giving the displacement of the particle of x = 0; (c) _ the function g (x) giving the shape of the string at t = 0; (d) plot the shape g (x) of the string at t = 0; and {e) plot the shape of the string of t = 5 s. [PoU 2021 Fall] Given, the wave equation in the CGS system is y = yy et? = (0.4 cm) elt!) 4 emi? y = 0.4 x ett-0253? (a) The velocity of the wave is v= & . & [0.4 x et-0258) or, v=-2 x (t-0.25 x) x 0.4 x e025"? v = -0.8 (t- 0.25 x) e075” (b) Atx = 0, the function giving the displacement of the particle is f() =04e" (c) Att = 0, the function g (x) gives the shape of the string is gi) =04e%" (d) The plot of the shape g (x) of the string at t = 0 is (e)