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1051-455-20073 Solution Set
- Determine which of the following describe traveling waves. Where appropriate, draw the profile and find the speed and direction of motion.
(a) ψ [y, t] = exp [− (a^2 y^2 + b^2 t^2 − 2 abty)] the argument of a traveling wave must have the form Φ [y, t] = k 0 z − ω 0 t where k 0 and ω 0 are constants. In this case
f [y, t] = exp
a^2 y^2 + b^2 t^2 − 2 abty
= exp
− (ay − bt)^2
= exp
− (Φ [y, t])^2
The argument u = ay − bt does have the proper form for the function f [u] = exp [−u^2 ], so this is a traveling wave. The argument remains constant for larger y if t increases, so the form moves to the right towards +∞. Examples are plotted below for different times t 0 and t 1 > t 0 for a = b = 1
-5 -4 -3 -2 -1 0 1 2 3 4 5
y
f[y,t]
f [y, t = 0] (black) and f [y, t = 1] (red) for a = b = 1
(b) ψ [z, t] = A sin [az^2 − bt^2 ]
we can check this one by applying it to the wave equation ∂ (^2) ψ ∂z^2 =^ 1 v^2
∂^2 ψ ∂t^2 :
=⇒ v^2 =
∂^2 ψ ∂t^2
∂^2 ψ ∂z^2
∂ψ ∂z
= A
∂z sin
az^2 − bt^2
= 2Aaz cos
bt^2 − az^2
∂^2 ψ ∂z^2
∂z
2 Aaz cos
bt^2 − az^2
= 2Aa cos
bt^2 − az^2
bt^2 − az^2
∂ψ ∂t
= A
∂t
sin
az^2 − bt^2
= − 2 Abt cos
bt^2 − az^2
∂^2 ψ ∂t^2
∂t
− 2 Abt cos
bt^2 − az^2
= 4Ab^2 t^2 sin
bt^2 − az^2
− 2 Ab cos
bt^2 − az^2
∂^2 ψ ∂t^2
∂^2 ψ ∂z^2
4 Ab^2 t^2 sin (bt^2 − az^2 ) − 2 Ab cos (bt^2 − az^2 ) 4 Aa^2 z^2 sin (bt^2 − az^2 ) + 2Aa cos (bt^2 − az^2 )
which does not reduce to a velocity, so NOT a traeveling wave. (c) ψ [x, t] = A sin
h 2 π
¡x a +^
t b
¢ 2 i
This has the proper form where the velocity is v = λ 0 ν 0 = ab. The argument shows that x must decrease as t increases, so the function moves to the left to- wards x = −∞. The function is shown below for A = 1 unit, a = 1 unit, and b = 1 sec for t 0 = 0 and t 1 = 1 :
-5 -4 -3 -2 -1 1 2 3 4 5
-1.
-0.
-0.
-0.
-0.
x
f[x,t]
f [y, t = 0] (black) and f [y, t = 1] (red) for A = a = b = 1 unit
- The figure shows the profile of a transverse wave on a string traveling in the positive z-direction at a speed of 1 ms.
(a) Determine its wavelength. the distance between adjacent maxima or minima is Z 0 = 300 mm (b) Notice that as the wave passes any fixed point on the z-axis, the string at that location oscillates in time. Draw a graph of ψ [t] showing how a point on the rope at z = 0 oscillates. since the velocity is 1 ms , then the function moves past the viewer at one location through 313 cycles in one second. The graph of the temporal behavior looks just like that of the spatial behavior except the horizontal axis becomes time in seconds and the unit of z = 300 mm becomes t = 103 sec. (c) What is the temporal frequency of the wave? As just shown, a wave with period λ 0 = 300 mm that moves at v = 1 ms has one cycle go past in 103 sec, which means that the frequency is 103 Hz
- A particle executing simple harmonic motion given by
y [t] = 4 sin
2 π t 6
is displaced by +1 unit when t = 0. Find:
(a) the phase angle Φ [t = 0] ≡ φ 0
y [0] = +1 = 4 sin
2 π
= 4 sin [φ 0 ]
=⇒ sin [φ 0 ] = +
=⇒ φ 0 = sin−^1
∼= 0. 252 radians ∼= 14. 4 ◦^ ∼= 14◦ 240
=⇒ y [t] = 4 sin
2 π t 6
∼= 4 sin
2 π t 6
(b) the difference in phase between any two positions of the particle separated in time by 2 sec; t 1 & t 2 are any two times such that t 2 − t 1 = 2 sec
y [t 2 ] = 4 sin
2 π t 2 6
y [t 1 ] = 4 sin
2 π t 1 6
∆Φ = Φ [t 2 ] − Φ [t 1 ] = 2π t 2 − t 1 6 = 2π 2 sec 6 sec
2 π 3 radians = 120◦
(c) the phase angle corresponding to a displacement of +2; if y = +2 units, then the initial equation is:
+2 = 4 sin
2 π t 6
= 4 sin [Φ [t]]
=⇒ sin [Φ [t]] =
=⇒ Φ [t] = π 6
(d) the time necessary to reach a displacement of +3 units from the initial position. if y = +3 units, then the initial equation is:
+3 = 4 sin
2 π t 6
= 4 sin [Φ [t]]
=⇒ sin [Φ [t]] =
=⇒ Φ [t] = sin−^1
=⇒ 2 π t 6
=⇒ 2 π t 6
∼= 34◦ 160 ∼= 0.598 =⇒ t ∼= 6 2 π · 0. 598 ∼= 0.57 sec
- By finding appropriate relations for k • r, write equations describing a sinusoidal plane wave in three directions in terms of wavelength and velocity for the three cases:
(a) propagation along the x-axis;
k =
2 π λ 0 0 0
⎦ (^) =⇒ ψ [r, t] = A 0 cos
2 π λ 0 (x − λ 0 ν 0 t)
= A 0 cos
2 π λ 0 (x − vφt)
(b) propagation along the line x = y; z = 0;
k =
2 π λ 0 cos^
£π 4
2 π λ 0 sin^
£π 4
√^2 π 22 πλ^0 √ 2 λ 0 0
⎦ (^) =⇒ ψ [r, t] = A 0 cos
2 π λ 0
μ x √ + y 2
± vφt
(c) propagation perpendicular to the planes x + y + z = k where k is a constant.
k =
kx ky kz
⎦ (^) where kx = ky = kz ≡ a =⇒ k =
a a a
⎦ (^) and |k| =^2 π λ 0
3 a
=⇒ a = √^2 π 3 λ 0 =⇒ ψ [r, t] = A 0 cos
2 π √ 3 λ 0
x + y + z ± vφt ·
- Two waves of the same amplitude, speed, and frequency travel together in the same region of space. The resultant wave may be written as a sum of two individual waves:
ψ [z, t] = A 0 sin [k 0 z + ω 0 t] + A 0 sin [k 0 z − ω 0 t + π]
With the help of complex exponentials, show that:
ψ [z, t] = 2A 0 cos [k 0 z] · sin [ω 0 t]
A 0 sin [k 0 z + ω 0 t] + A 0 sin [k 0 z − ω 0 t + π] = A 0 sin [k 0 z + ω 0 t] − A 0 sin [k 0 z − ω 0 t] = A 0 Im {exp [+i (k 0 z + ω 0 t)]} − A 0 Im {exp [+i (k 0 z − ω 0 t)]}
A 0 Im {exp [+ik 0 z] (exp [+iω 0 t] − exp [−iω 0 t])} = A 0 Im {exp [+ik 0 z] · 2 i sin [+ω 0 t]} = 2 A 0 sin [+ω 0 t] Im^ {i^ ·^ exp [+ik 0 z]} = 2 A 0 sin [+ω 0 t] Im {i · (cos [+k 0 z] + i sin [+k 0 z])} = 2 A 0 sin [+ω 0 t] Im {(i · cos [+k 0 z] − sin [+k 0 z])} = 2 A 0 cos [k 0 z] · sin [ω 0 t]
