1. Sinusoidal Waves
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

2. uy(x,t) =  y/ t = “partial derivative with respect to t”
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”

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/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?

5. Wave speed of a stretched string
v = C (F/)1/ (MLT-2/ML-1)1/2 =L/T
detailed calculation using 2nd law yields C= 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)= 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

13. 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

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