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Sinusoidal Waves

Sinusoidal Waves. y(x,t)=y m 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)=y m sin( kx +  t) is a wave moving left. Transverse Velocity.

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Sinusoidal Waves

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  1. Sinusoidal Waves • 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

  2. 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”

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

  4. 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/lengthML-1 • tension F is the elastic character (a force) MLT-2 • how can we combine tension and mass density to get units of speed?

  5. 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

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

  7. 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

  8. Waves F F F/2 F/2 v=(F/)1/2

  9. 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

  10. 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

  11. 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

  12. Energy and Power • y(x,t)= ymsin( kx- t) • uy=dy/dt= -ym cos( kx- t) (keep x fixed!) • dK=(1/2)dm uy2 =(1/2) dx 2ym2cos2(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

  13. Potential Energy • Length • hence dl-dx = (1/2) (dy/dx)2 dx • dU = (1/2) F (dy/dx)2 dxpotential energy of element dx • y(x,t)= ymsin( kx- t) • dy/dx= ym k cos(kx - t) keeping t fixed! • Since F=v2 = 2/k2 we find • dU=(1/2) dx 2ym2cos2(kx-  t) • dK=(1/2) dx  2ym2cos2(kx-  t) • dE= 2ym2cos2(kx- t) dx • average of cos2 over one period is 1/2 • dEav= (1/2)   2ym2 dx

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

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