Sinusoidal Waves 16.107 LO1 Jan 14/02 y(x,t)=ym sin( kx-  t) describes a wave moving right at constant speed v= /k  = 2f = 2/T k= 2/ v = /k = f = /T wave speed= one wavelength per period y(x,t)=ym sin( kx+ t) is a wave moving left

uy(x,t) =  y/ t = “partial derivative with respect to t” 16.107 LO1 Jan 14/02 Transverse Velocity y(x,t)=ym sin( kx- t) uy(x,t) =  y/ t = “partial derivative with respect to t” “derivative of y with respect to ‘t’ keeping ‘x’ fixed” = -ym  cos( kx- t) maximum transverse speed is ym  A more general form is y(x,t)=ym sin( kx- t-) (kx- t-) is the phase of the wave two waves with the same phase or phases differing by 2n are said to be “in phase”

Phase and Phase Constant y(x,t)=ym sin( kx- t-) =ym sin[ k(x -/k) - t] =ym sin[ kx- (t+/)]

Wave speed of a stretched string Actual value of v = /k is determined by the medium as wave passes, the “particles” in the medium oscillate medium has both inertia (KE) and elasticity (PE) dimensional argument: v= length/time LT-1 inertia is the mass of an element =mass/length ML-1 tension F is the elastic character (a force) MLT-2 how can we combine tension and mass density to get units of speed?

Wave speed of a stretched string v = C (F/)1/2 (MLT-2/ML-1)1/2 =L/T detailed calculation using 2nd law yields C=1 v = (F/)1/2 speed depends only on characteristics of string independent of the frequency of the wave f due to source that produced it once f is determined by the generator, then  = v/f = vT

(a) 2,3,1 (b) 3,(1,2)

Summary  = 2f = 2/T k= 2/ v = /k = f = /T wave speed= one wavelength per period y(x,t)=ym sin( kx- t-) describes a wave moving right at constant speed v=  /k y(x,t)=ym sin( kx+ t-) is a wave moving left v = (F/)1/2 F = tension  = mass per unit length

Waves F F F/2 F/2 v=(F/)1/2

Wave Equation How are derivatives of y(x,t) with respect to both x and t related => wave equation length of segment is x and its mass is m= x net force in vertical direction is Fsin2 - Fsin 1 but sin~ ~tan  when  is small net vertical force on segment is F(tan2 - tan 1 ) but slope S of string is S=tan  = y/x net force is F(S2 - S1) = F S = ma = x2y/t2

Wave Equation F S = x2y/t2 force = ma S/x = (/F)2y/t2 as x => 0, S/x = S/ x = / x (y/ x)= 2y/x2 any function y=f(x-vt) or y=g(x+vt) satisfies this equation with v = (F/)1/2 y(x,t)= A sin(kx-t) is a harmonic wave

Energy and Power it takes energy to set up a wave on a stretched string y(x,t)=ym sin( kx- t) the wave transports the energy both as kinetic energy and elastic potential energy an element of length dx of the string has mass dm = dx this element (at some pt x) moves up and down with varying velocity u = dy/dt (keep x fixed!) this element has kinetic energy dK=(1/2)(dm)u2 u is maximum as element moves through y=0 u is zero when y=ym

Energy and Power y(x,t)= ym sin( kx- t) uy=dy/dt= -ym  cos( kx- t) (keep x fixed!) dK=(1/2)dm uy2 =(1/2) dx 2 ym2cos2(kx- t) kinetic energy of element dx potential energy of a segment is work done in stretching string and depends on the slope dy/dx when y=A the element has its normal length dx when y=0, the slope dy/dx is largest and the stretching is maximum dU = F( dl -dx) force times change in length both KE and PE are maximum when y=0

Potential Energy Length hence dl-dx = (1/2) (dy/dx)2 dx dU = (1/2) F (dy/dx)2 dx potential energy of element dx y(x,t)= ym sin( kx- t) dy/dx= ym k cos(kx -  t) keeping t fixed! Since F=v2 = 2/k2 we find dU=(1/2) dx 2ym 2cos2(kx-  t) dK=(1/2) dx  2ym 2cos2(kx-  t) dE= 2ym 2cos2(kx- t) dx average of cos2 over one period is 1/2 dEav= (1/2)   2ym 2 dx

cos(x) 0. cos2(x) .5