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Sinusoidal Waves
• y(x,t)=ym sin( kx- t) describes a wave moving right at constant speed v= /k
= 2f = 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
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
2n are said to be “in phase”
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
Summary = 2f = 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
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 Fsin2 - Fsin 1
• but sin~ ~tan when is small
• net vertical force on segment is F(tan2 - tan 1 )
• but slope S of string is S=tan = y/x
• net force is F(S2 - S1) = F S = ma = x2y/t2
2
2 2
2
2
1y
x
y
tv
Wave Equation• F S = x2y/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
2
2
2
2
y
x
y
t
F
2
2 2
2
2
1y
x
y
tv
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
d l dx dy dx dy dx dx dy dx dx 2 2 2 21 1 2( / ) ( / )( / )