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Physics 315: Oscillations and Waves
Midterm 2. Solutions.
- Successive nodes (N) in a standing wave pattern are λ/2 apart, where λ is the wavelength.
The anti-nodes (A) lie halfway between the nodes. The dispersion relation for transverse
waves of phase velocity v is
v = f λ, (1)
where f is the frequency (in Hertz).
(a) Both ends of the string are stationary, and, therefore, correspond to nodes. Thus,
the length, l, of the string is either λ/2 (corresponding to NAN), λ (corresponding
to NANAN), 3λ/2 (corresponding to NANANAN), et cetera. In other words,
l =
j λ
where j = 1 , 2, 3, et cetera. It follows that
λ (^) j =
2 l
j
and
f (^) j =
j v
2 l
Hence, the wavelength and frequency of the longest wavelength mode (i.e., j = 1) are
λ 1 = 2 l, f 1 =
v
2 l
The wavelength and frequency of the next longest wavelength mode (i.e., j = 2) is
λ 2 = l, f 2 =
v
l
(b) The most general expression for y(x, t) is a linear superposition of the normal modes of
oscillation: that is,
y(x, t) =
j= 1 ,∞
c (^) j sin
2 π
x
λ (^) j
cos
2 π f (^) j t − φ (^) j
where the c (^) j and φ (^) j are arbitrary constants.
(c) It follows from the previous expression that
∂y(x, t)
∂t
j= 1 ,∞
2 π f (^) j c (^) j sin
2 π
x
λ (^) j
sin
2 π f (^) j t − φ (^) j
In particular,
∂y(x, 0)
∂t
j= 1 ,∞
2 π f (^) j c (^) j sin
2 π
x
λ (^) j
sin φ (^) j. (9)
We can ensure that ∂y(x, 0)/∂t = 0 by setting φ (^) j = 0 for all j. We are left with
y(x, 0) =
j= 1 ,∞
c (^) j sin
2 π
x
λ (^) j
The given initial condition can be written
y(x, 0) = a sin
2 π
x
λ 1
2 π
x
λ 3
Hence, it is clear that c 1 = a, c 2 = 0, c 3 = b, and c (^) j> 3 = 0. Thus, the motion of the
string is specified by
y(x, t) = a sin
2 π
x
λ 1
cos (2π f 1 t)^ + b sin
2 π
x
λ 3
cos (2π f 3 t)
= a sin
π
x
l
cos
π
v t
l
3 π
x
l
cos
3 π
v t
l
- (a) The first term on the right-hand side of the waveform
V(x, t) = V (^) i cos
[(
ω
c
(x − c t)
]
[(
ω
c
(x + c t)
]
corresponds to an incident traveling wave of amplitude V (^) i and angular frequency ω,
propagating in the +x direction at the phase velocity c. The second term corresponds
to a reflected traveling wave of amplitude V (^) r and angular frequency ω, propagating in
the −x direction at the phase velocity c. Given that
∂I
∂t
c
Z 1
∂V
∂x
we have ∂I
∂t
ω V (^) i
Z 1
sin
[(
ω
c
(x − c t)
]
ω V (^) r
Z 1
sin
[(
ω
c
(x + c t)
]
Integration with respect to t yields
I(x, t) =
V (^) i
Z 1
cos
[(
ω
c
(x − c t)
]
V (^) r
Z 1
cos
[(
ω
c
(x + c t)
]
(c) Continuity of V(x, t) at x = 0 yields
V (^) i cos(ω t) + V (^) r cos(ω t) = V (^) t cos(ω t), (25)
or
V (^) i + V (^) r = V (^) t. (26)
Continuity of the energy flux gives
V
2 i
− V
2 r
2 Z 1
V
2 t
2 Z 2
Equations (26) and (27) can be combined to give
V (^) i − V (^) r =
Z 1
Z 2
V (^) t. (28)
Equations (26) and (28) can be solved to give
V (^) t =
2 Z 2
Z 1 + Z 2
V (^) i, (29)
V (^) r =
Z 1 − Z 2
Z 1 + Z 2
V (^) i. (30)
Thus, the coefficient of transmission is
T ≡
〈It〉
〈Ii〉
Z 1
Z 2
V
2 t
V
2 i
4 Z 1 Z 2
(Z 1 + Z 2 )
2
