451
16-1 TRANSVE RSE WAVES
crease (mentally slide the curve slightly rightward). If,
instead, the wave is moving leftward, then just after the snapshot, the depth at x ϭ 0 should decrease. Now let’s check the
graph in Fig. 16-8b. It tells us that just after t ϭ 0, the depth increases.Thus, the wave is moving rightward, in the positive direction of x, and we choose the minus sign in Eq. 16-18.
Phase constant: The value of f is set by the conditions at
x ϭ 0 at the instant t ϭ 0. From either figure we see that at
that location and time, y ϭ Ϫ2.0 mm. Substituting these
three values and also y m ϭ 3.0 mm into Eq. 16-18 gives us
Ϫ2.0 mm ϭ (3.0 mm) sin(0 ϩ 0 ϩ f)
or
f ϭ sin
Ϫ1
ϭ Ϫ0.73 rad.
Note that this is consistent with the rule that on a plot of y
versus x, a negative phase constant shifts the normal sine
function rightward, which is what we see in Fig. 16-8a.
Equation: Now we can fill out Eq. 16-18:
y ϭ (3.0 mm) sin(200px Ϫ l00pt Ϫ 0.73 rad), (Answer)
with x in meters and t in seconds.
(Ϫ
2
3 )
paper sheet on the graph, mark those crossing points, slide
the sheet to align the left-hand mark with the origin, and
then read off the location of the right-hand mark. Either
way we find l ϭ 10 mm. From Eq. 16-5, we then have
Period: The period T is the time interval that a string element’s SHM takes to begin repeating itself. In Fig. 16-8b, T
is the distance along the t axis from one crossing point to the
next crossing point where the plot has the same slope.
Measuring the distance visually or with the aid of a sheet of
paper, we find T ϭ 20 ms. From Eq. 16-8, we then have
Direction of travel: To find the direction, we apply a bit of
reasoning to the figures. In the snapshot at t ϭ 0 given in
Fig. 16-8a, note that if the wave is moving rightward, then just
after the snapshot, the depth of the wave at x ϭ 0 should inv ϭ
2p
T
ϭ
2p
0.020 s
ϭ 100p rad/s.
k ϭ
2p
l
ϭ
2p
0.010 m
ϭ 200p rad/m.
Next, substituting numerical values but suppressing the
units, which are SI, we write
u ϭ (Ϫ2.72)(0.00327) cos[(72.1)(0.225) Ϫ (2.72)(18.9)]
ϭ 0.00720 m/s ϭ 7.20 mm/s.
(Answer)
Thus, at t ϭ 18.9 s our string element is moving in the
positive direction of y with a speed of 7.20 mm/s.
(Caution: In evaluating the cosine function, we keep all the
significant figures in the argument or the calculation can be
off considerably. For example, round off the numbers to two
significant figures and then see what you get for u.)
(b) What is the transverse acceleration a y of our string
element at t ϭ 18.9 s?
KEY IDEA
The transverse acceleration a y is the rate at which the element’s transverse velocity is changing.
Calculations: From Eq. 16-20, again treating x as a constant
but allowing t to vary, we find
(16-21)
Substituting numerical values but suppressing the units,
which are SI, we have
a y 5 2(2.72)
2
(0.00327) sin[(72.1)(0.225) Ϫ (2.72)(18.9)]
ϭ Ϫ0.0142 m/s
2 ϭ Ϫ14.2 mm/s
2
.
( A n s w e r )
a y ϭ
Ѩu
Ѩt
ϭ Ϫv
2
y m sin (kx Ϫ vt).
Sample Problem 16.02 Transverse velocity and transverse acceleration of a string element
A wave traveling along a string is described by
y(x, t) = (0.00327 m) sin(72.1x Ϫ 2.72t),
in which the numerical constants are in SI units (72.1 rad/m
and 2.72 rad/s).
(a) What is the transverse velocity u of the string element
at x ϭ 22.5 cm at time t ϭ 18.9 s? (This velocity, which is
associated with the transverse oscillation of a string
element, is parallel to the y axis. Don’t confuse it with v,
the constant velocity at which the wave form moves along
the x axis.)
KEY IDEAS
The transverse velocity u is the rate at which the
displacement y of the element is changing. In general, that
displacement is given by
y(x, t) = y m sin(kx Ϫ vt).
(16-19)
For an element at a certain location x, we find the rate of
change of y by taking the derivative of Eq. 16-19 with respect to t while treating x as a constant. A derivative taken
while one (or more) of the variables is treated as a constant
is called a partial derivative and is represented by a symbol
such as
rather than d/dt.
Calculations: Here we have
(16-20)
u ϭ
Ѩy
Ѩt
ϭ Ϫvy m cos(kx Ϫ vt).
Ѩ/Ѩt
16-1 TRANSVE RSE WAVES
crease (mentally slide the curve slightly rightward). If,
instead, the wave is moving leftward, then just after the snapshot, the depth at x ϭ 0 should decrease. Now let’s check the
graph in Fig. 16-8b. It tells us that just after t ϭ 0, the depth increases.Thus, the wave is moving rightward, in the positive direction of x, and we choose the minus sign in Eq. 16-18.
Phase constant: The value of f is set by the conditions at
x ϭ 0 at the instant t ϭ 0. From either figure we see that at
that location and time, y ϭ Ϫ2.0 mm. Substituting these
three values and also y m ϭ 3.0 mm into Eq. 16-18 gives us
Ϫ2.0 mm ϭ (3.0 mm) sin(0 ϩ 0 ϩ f)
or
f ϭ sin
Ϫ1
ϭ Ϫ0.73 rad.
Note that this is consistent with the rule that on a plot of y
versus x, a negative phase constant shifts the normal sine
function rightward, which is what we see in Fig. 16-8a.
Equation: Now we can fill out Eq. 16-18:
y ϭ (3.0 mm) sin(200px Ϫ l00pt Ϫ 0.73 rad), (Answer)
with x in meters and t in seconds.
(Ϫ
2
3 )
paper sheet on the graph, mark those crossing points, slide
the sheet to align the left-hand mark with the origin, and
then read off the location of the right-hand mark. Either
way we find l ϭ 10 mm. From Eq. 16-5, we then have
Period: The period T is the time interval that a string element’s SHM takes to begin repeating itself. In Fig. 16-8b, T
is the distance along the t axis from one crossing point to the
next crossing point where the plot has the same slope.
Measuring the distance visually or with the aid of a sheet of
paper, we find T ϭ 20 ms. From Eq. 16-8, we then have
Direction of travel: To find the direction, we apply a bit of
reasoning to the figures. In the snapshot at t ϭ 0 given in
Fig. 16-8a, note that if the wave is moving rightward, then just
after the snapshot, the depth of the wave at x ϭ 0 should inv ϭ
2p
T
ϭ
2p
0.020 s
ϭ 100p rad/s.
k ϭ
2p
l
ϭ
2p
0.010 m
ϭ 200p rad/m.
Next, substituting numerical values but suppressing the
units, which are SI, we write
u ϭ (Ϫ2.72)(0.00327) cos[(72.1)(0.225) Ϫ (2.72)(18.9)]
ϭ 0.00720 m/s ϭ 7.20 mm/s.
(Answer)
Thus, at t ϭ 18.9 s our string element is moving in the
positive direction of y with a speed of 7.20 mm/s.
(Caution: In evaluating the cosine function, we keep all the
significant figures in the argument or the calculation can be
off considerably. For example, round off the numbers to two
significant figures and then see what you get for u.)
(b) What is the transverse acceleration a y of our string
element at t ϭ 18.9 s?
KEY IDEA
The transverse acceleration a y is the rate at which the element’s transverse velocity is changing.
Calculations: From Eq. 16-20, again treating x as a constant
but allowing t to vary, we find
(16-21)
Substituting numerical values but suppressing the units,
which are SI, we have
a y 5 2(2.72)
2
(0.00327) sin[(72.1)(0.225) Ϫ (2.72)(18.9)]
ϭ Ϫ0.0142 m/s
2 ϭ Ϫ14.2 mm/s
2
.
( A n s w e r )
a y ϭ
Ѩu
Ѩt
ϭ Ϫv
2
y m sin (kx Ϫ vt).
Sample Problem 16.02 Transverse velocity and transverse acceleration of a string element
A wave traveling along a string is described by
y(x, t) = (0.00327 m) sin(72.1x Ϫ 2.72t),
in which the numerical constants are in SI units (72.1 rad/m
and 2.72 rad/s).
(a) What is the transverse velocity u of the string element
at x ϭ 22.5 cm at time t ϭ 18.9 s? (This velocity, which is
associated with the transverse oscillation of a string
element, is parallel to the y axis. Don’t confuse it with v,
the constant velocity at which the wave form moves along
the x axis.)
KEY IDEAS
The transverse velocity u is the rate at which the
displacement y of the element is changing. In general, that
displacement is given by
y(x, t) = y m sin(kx Ϫ vt).
(16-19)
For an element at a certain location x, we find the rate of
change of y by taking the derivative of Eq. 16-19 with respect to t while treating x as a constant. A derivative taken
while one (or more) of the variables is treated as a constant
is called a partial derivative and is represented by a symbol
such as
rather than d/dt.
Calculations: Here we have
(16-20)
u ϭ
Ѩy
Ѩt
ϭ Ϫvy m cos(kx Ϫ vt).
Ѩ/Ѩt