2. Transverse velocity of a particle Note that v is the speed of wave transmitting What is the velocity of particle oscillating? It is called transverse velocity of a particle for transverse wave Transverse velocity oy a l1,(x,) Lym sin( kx-at ) at at ym@ cos(kx-@t) (18-14) Tansverse acceleration a,(x, t) dr2=-ym @sin( kx-at=-o y (18-15)
2. Transverse velocity of a particle Note that is the speed of wave transmitting. v What is the velocity of particle oscillating? ---- It is called transverse velocity of a particle for transverse wave cos( ) ( , ) [ sin( )] y k x t y k x t t t y u x t m y m = − − − = = Transverse velocity: Tansverse acceleration: y k x t y dt d y a x t y m 2 2 2 2 ( , ) = = − sin( − ) = − (18-14) (18-15)
3 Phase and phase constant (18-16) If the equation of the wave is: y(, t)=ym sin(kx-ot-o) (kx-at-oPhase phase constant Eg(18-16)can be written in two equivalent forms y(x)= ym sin k(x-2)-](18-17 k 7a) y(x,1)=ynsi[kx-O(t+)(18-17b)
3. Phase and phase constant y(x,t) = y sin( k x−t −) m (kx −t −) If the equation of the wave is: Phase phase constant Eq(18-16) can be written in two equivalent forms: (18-17a) (18-17b) ( , ) sin[ ( ) t] k y x t y k x m = − − ( , ) sin[ ( )] y x t = ym k x− t + (18-16)
Two waves a and B: y=ym sin(h -- wave Alead y=ymsin(h-at) wave B lag e In y-x, wave a is ahead of wave b by a distance o/k v(x,t)=ym sin[ k(x-1)-] BXA b In y-t, wave A is ahead of e wave b by a time /w y(x,t)=ym sin[ kx-Q(t+RIAXB Fig 18-7
In y-x, wave A is ahead of wave B by a distance /k In y-t, wave A is ahead of wave B by a time /ω (a) x k B A y y = ymsin(kx – ωt – ) Two waves A and B: y = ymsin(kx – ωt ) wave A wave B (b) t Fig 18-7 y A B ( , ) sin[ ( ) t] k y x t y k x m = − − ( , ) sin[ ( )] y x t = ym k x− t + lead lag
Sample problem 18-1 A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance of 1. 30cm The motion is repeated regularly 125 times per second (a If the distance between adjacent wave crests iS 15.6 cm, find the amplitude, frequency, speed and wavelength of the wave (b) Assuming the wave moves in the +X direction e and that at t=0, the element of the string at X=0 is at its equilibrium position y=o and moving downward, find the equation of the wave
Sample problem 18-1 A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance of 1.30cm. The motion is repeated regularly 125 times per second (a) If the distance between adjacent wave crests is 15.6 cm, find the amplitude, frequency, speed, and wavelength of the wave . (b) Assuming the wave moves in the +x direction and that at t=0, the element of the string at x=0 is at its equilibrium position y=0 and moving downward, find the equation of the wave
Solution 1.30 (a the amplitude ym.=cm=0.65cm frequency f=125Hz wavelength =156cm speed f=19.5m/S (b) The general expression for a sinusoidal waves is given by Eq(18-16 y(, t=ym sin( kx-ot-o)
Solution: (a) The amplitude frequency wavelength speed (b) The general expression for a sinusoidal waves is given by Eq(18-16) ym cm 0.65cm 2 1.30 = = f = 125Hz v = f = 19.5m/s =15.6cm y(x,t) = y sin( k x−t −) m