16.107 LO1 Jan 14/02
Sinusoidal Waves
• y(x,t)=ym sin( kx- ω t) describes a wave
moving right at constant speed v= ω/k
• ω = 2πf = 2π/T
k= 2π/λ
• v = ω/k = f λ= λ/T
• wave speed= one wavelength per period
• y(x,t)=ym sin( kx+ ωt) is a wave moving left
Phase and Phase Constant
• y(x,t)=ym sin( kx- ωt-φ) =ym sin[ k(x -φ/k) - ωt] =ym sin[ kx- ω(t+φ/ω)]
Transverse Velocity
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y(x,t)=ym sin( kx- ωt)
uy(x,t) = ∂ y/∂ t = “partial derivative with respect to t”
“derivative of y with respect to ‘t’ keeping ‘x’ fixed”
= -ym ω cos( kx- ωt)
maximum transverse speed is ym ω
• A more general form is y(x,t)=ym sin( kx- ωt-φ)
• (kx- ωt-φ) is the phase of the wave
• two waves with the same phase or phases differing by
2πn are said to be “in phase”
Wave speed of a stretched string
• Actual value of v = ω/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 µ=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/µ)1/2
(MLT-2/ML-1)1/2 =L/T
• detailed calculation using 2nd law yields C=1
v = (F/µ)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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λ = v/f = vT
(a) 2,3,1
(b) 3,(1,2)
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16.107 LO1 Jan 14/02
Summary
• ω = 2πf = 2π/T
k= 2π/λ
•
v = ω/k = f λ= λ/T
• wave speed= one wavelength per period
Waves
• y(x,t)=ym sin( kx- ωt-φ) describes a wave
moving right at constant speed v= ω /k
• y(x,t)=ym sin( kx+ ωt-φ) is a wave moving
left
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v = (F/µ)1/2
• F = tension
µ = mass per unit length
F
F
F/2
F/2
v=(F/µ)1/2
Wave Equation
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=µ ∆x
net force in vertical direction is Fsinθ2 - Fsin θ1
but sinθ~ θ~tan θ when θ is small
net vertical force on segment is F(tanθ2 - tan θ1 )
but slope S of string is S=tan θ = ∂y/∂x
net force is F(S2 - S1) = F ∆S = ma = µ∆x∂2y/∂t2
• F ∆S = µ∆x∂2y/∂t2
• ∆S/∆x = (µ/F)∂2y/∂t2
• as ∆x => 0, ∆S/∆x = ∂S/ ∂x = ∂/ ∂x (∂y/ ∂x)= ∂2y/∂x2
∂2 y µ ∂2 y
=
∂x 2 F ∂t 2
• any function y=f(x-vt) or y=g(x+vt) satisfies this equation
with
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Energy and Power
• it takes energy to set up a wave on a stretched
string
y(x,t)=ym sin( kx- ωt)
• the wave transports the energy both as kinetic
energy and elastic potential energy
• an element of length dx of the string has mass
dm = µdx
• 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
force = ma
v = (F/µ)1/2
∂2 y 1 ∂2 y
=
∂x 2 v 2 ∂t 2
y(x,t)= A sin(kx-ωt) is a harmonic wave
Energy and Power
• y(x,t)= ym sin( kx- ωt)
• uy=dy/dt= -ym ω cos( kx- ωt) (keep x fixed!)
• dK=(1/2)dm uy2 =(1/2) µdx ω2 ym2cos2(kx- ωt)
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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
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16.107 LO1 Jan 14/02
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- ωt)
keeping t fixed!
dy/dx= ym k cos(kx - ω t)
Since F=µv2 = µω2/k2 we find
dU=(1/2) µdx ω2ym 2cos2(kx- ω t)
dK=(1/2) µdx ω 2ym 2cos2(kx- ω t)
dE= µω2ym 2cos2(kx- ωt) dx
average of cos2 over one period is 1/2
dEav= (1/2) µ ω 2ym 2 dx
cos(x)
cos2(x)
0.
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