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”

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 = 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

Waves

v=(F/)1/2

F F/2 F/2

F

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( / ) ( / )( / )

.5

0.cos(x)

cos2(x)