Download Energy and Power

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
Transverse Velocity
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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
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 = v/f = vT
(a) 2,3,1
(b) 3,(1,2)
Summary
•  = 2f = 2/T
k= 2/
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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
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v = (F/)1/2
• F = tension
 = mass per unit length
Waves
F
F
v=(F/)1/2
F/2
F/2
Wave Equation
• How are derivatives of y(x,t) with respect to both x and t
related => wave equation
2 y 1 2 y
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x
2

v 2 t 2
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
2 y  2 y

2
x
F t 2
• any function y=f(x-vt) or y=g(x+vt) satisfies this equation
with
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v = (F/)1/2
2 y 1 2 y
 2 2
2
x
v t
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)
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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 dl  dx 2  dy 2  dx 1  (dy / dx)2  dx  (1 / 2)(dy / dx)2 dx
• hence dl-dx = (1/2) (dy/dx)2 dx
• dU = (1/2) F (dy/dx)2 dx potential energy of element dx
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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)
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