 # Recitation 2

```Recitation 2
Chapters 12 and 13
Problem 12.20. A 65.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her and the bridge (Figure
P12.20). The unstretched length of the cord is 11.0 m. She reaches the bottom of her motion 36.0 m below the bridge before
bouncing back. Her motion can be seperated into an 11.0 m free-fall and a 25.0 m section of simple harmonic oscillation. (a)
For what time interval is she in free-fall? (b) Use the principle of consevation of energy to find the spring constant of the
bungee cord. (c) What is the location of the equilibrium point where the spring force balances the gravitational force acting
on the jumper? Note that this point is takes as the origin in our mathematical description of simple harmonic oscillation.
(d) What is the angular frequency of the oscillation? (e) What time interval is required for the cord to stretch by 25.0 m? (f )
What is the total time interval for the entire 36.0 m drop?
It always helps me to draw a picture of what’s going on:
The dotted horizontal lines are, top to bottom, the position where the bungee cord begins resisting the fall and the equilibrium
position of the simple harmonic oscillation. The dotted vertical lines are, left to right, the times of the person jumping off the
bridge, the bungee cord beginning to resist the fall, and the bungee cord ceasing to resist the fall as the jumper is launched
back up into the air. The red curve tracks the position of the jumper as a function of time, and the blue curve extrapolates
the simple harmonic oscillation to draw attention to the difference between the simple harmonic oscillation and free-fall
portions of the actual trajectory.
(a) The free-fall phase follows the parabolic y = − 21 gt2 behavior you all know and love from your Freshman Mechanics class.
This fall continues until the jumper drops the L = 11 m needed to take the slack out of the bungee cord. The time is thus
s
1 2
2L
−L = − gtf
tf =
= 1.50 s .
(1)
2
g
(b) We don’t know anything about velocity yet, so lets conserve energy between the points where the velocity is zero. Defining
hb = 0 to be the height of the lowest point in the trajectory, the energy at that point is all spring potential energy. Of course,
the energy at the jumping-off point (ht = 36 m)is all gravitational potential energy. Setting these equal, we have
1
1
2
k∆L2 = k (ht − L)
2
2
E = mght =
k=
2mght
2
(ht − L)
(2)
= 73.4 N/m
(3)
where ∆L = ht − L = 25 m is the maximum stretch in the bungee cord. Of course, in the real world energy is lost to heating
the bungee cord, shaking the jumper, etc., which is good, since otherwise the jumper would bump into the bridge in the
second free-fall phase on the right side of the drawing above.
(c) Here we just use Hooke’s law and balance the vertical forces on the jumper.
mg = k∆Leq
∆Leq =
mg
= 8.68 m ,
k
(4)
where ∆Leq is the distance from bungee-engage to equilibrium. The equilibrium point is thus
L + ∆Leq = 19.7 m
(5)
below the bridge.
(d)
r
ω=
k
m
(e) Let’s take a look at the reference circle for the harmonic oscillation phase
(6)
∆Leq θ
A
so
θ = arcsin
ts =
θ+
ω
π
2
∆Leq
A
(7)
= 2.00 s .
(8)
(f) Combining the free-fall time from (a) with the harmonic time from (e) we have
ttb = tf + ts = 3.50 s .
(9)
Problem 12.42. (a) A hanging spring stretches by 35.0 cm when an object of mass 450 g is hung on it at rest. In this
situation, we define its position as x = 0. The object is pulled down an additional 18.0 cm and released from rest to oscillate
without friction. What is its position x at a time 84.4 s later? (b) A hanging spring stretches by 35.5 cm when an object of
mass 440 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 18.0 cm
and released from rest to oscillate without friction. Find its position 84.4 s later. (c) Why are the answers to parts (a) and
(b) different by such a large percentage when the data are so similar? Does this circumstance reveal a fundamental difficulty
in calculating the future? (d) Find the distance traveled by the vibrating object in (a). (e) Find the distance traveled by the
object in (b).
There’s a lot of repetition in this problem, so let’s do everything symbolically first. Ap
stretch ofp∆x due to hanging a mass
m implies a spring constant of k = F/∆x = mg/∆x and an angular frequency of ω = k/m = g/∆x. Notice that ω does
not depend on the hanging mass. At this point we know everything about how the position changes as a function of time,
namely
p
x(t) = −A cos(ωt) = −A cos( g/∆xt) ,
(10)
where the − cos part came from letting up be the positive x direction and noticing that the mass starts at an extreme low
point in its oscillation. Finally, we note that the total distance traveled as a function of time is going to be
d = 2A · NT /2 + (A + x(tfrac ))
(11)
where NT /2 = f loor(2t/T ) is the number of completed half-periods and tfrac = t − NT /2 · T /2 is the fractional half-period
left over. The extra A in the right-hand term ensures that the right-hand term is zero when tfrac = 0. Now we just have to
plug in the two cases. . .
(a)
∆x = 0.350 m
A = 0.180 m
t = 84.4 s
x(t) = −15.8 cm
(12)
(13)
(b)
∆x = 0.355 m
A = 0.180 m
x(t) = 15.9 cm
t = 84.4 s
(14)
(15)
(c) The answers to (a) and (b) are so different because the angular velocities are slightly different. The phase starts out
the same but grows slightly faster in (a). The difference seems so large because the phase loops around at 2π, so a small
difference in the “total phase” ωt can produce a large difference in the “effective phase” ωt mod 2π.
This sensisivity to the initial conditions (here to ∆x) is the same sort of effect that gives rise to chaotic behavior (see
http://en.wikipedia.org/wiki/Chaos_theory#Chaotic_dynamics). The simple harmonic oscillator is not chaotic though,
because ∆x is not a dimension that changes with time. This means that predicting the future for a simple harmonic oscillator
is not that difficult. You may have to keep adjusting your model parameters as you get more data, but with time your model
will get better and better.
Compare this system to one that is chaotic, e.g. the damped-forced pendulum. In this situations, the sensitivity to initial
conditions also involves the changing parameters, like the position of the pendulum. Not only does your model have to be
perfect, but your understanding of where the pendulum is now must also be perfect. This is a much more difficult problem,
since the position of the pendulum is changing with time, so you can’t just wait and aquire a arbitarily precise time average.
This is why it is so hard to make long term predictions for chaotic systems.
(d)
2π
= 1.19 s
ω
d = 51.1 m + 2.17 cm = 51.1 m
NT /2 = 142
T =
tfrac = 93.9 ms
(16)
(17)
(e)
2π
= 1.20 s
ω
d = 50.8 m + 2.05 cm = 50.8 m
T =
NT /2 = 141
tfrac = 91.7 ms
(18)
(19)
Notice that the distance d doesn’t loop back on itself like the position x, so the small relative difference in “total phase”
leads to a small relative difference in distance between (d) and (e).
Problem 13.5. The wave function for a traveling wave on a taut string is (in SI units)
y(x, t) = (0.350 m) sin(10πt − 3πx + π/4) .
(20)
(a) What are the speed and direction of travel of the wave? (b) What is the vertical position of an element of the string at
t = 0, x = 0.100 m? (c) What are the wavelength and frequency of the wave? (d) What is the maximum transverse speed of
an element of the string?
(a) When t = 0 and x = 0, the phase of y is π/4. If we look some small time ∆t later, we’ll need some small positive
displacement ∆x = 10∆t/3 to find that same phase. So the direction is to the right (positive x), and the speed is
v=
10
∆x
=
= 3.33 m/s
∆t
3
(21)
(b) Plugging in
y(x = 0.100 m, t = 0) = 0.350 m · sin(−0.3π + π/4) = −5.48 cm
(22)
(c) To change the phase of the wave by 2π, we need either
λ = ∆x2π =
2
= 0.667 m
3
(23)
T = ∆t2π =
2
= 0.200 s
10
(24)
or
f=
1
= 5 Hz
T
(25)
(d) We can find the speed at any time and position by differentiating y(x, t) with respect to time. After last week’s recitation,
we have lots of practice with such derivatives, so we see right off that the only effect on the amplitude will be to chain-rule
out a copy of the time coefficient ω = 10π rad/s, so
∂y
= Aω = 0.350 m · 10π 1/s = 11.0 m/s
(26)
∂t max
Problem 13.7. The string shown in Active Figure 13.8 is driven at a frequency of 5.00 Hz. The amplitude of the motion
is 12.0 cm and the wave speed is 20.0 m/s. Furthermore, the wave is such that y = 0 at x = 0 and t = 0. Determine (a)
the angular frequency and (b) wave number for this wave. (c) Write an expression for the wave function. Calculate (d) the
maximum transverse speed and (e) the maximum transverse acceleration of an element of the string.
(a) This is just a unit conversion.
=
·
s
cycle
s
ω = 2πf = 31.4 rad/s
(27)
(28)
(b) Another units conversion
=
·
m
s m
ω
v
(29)
(30)
(c) Now we get to plug in those values for k and ω.
y(x, t) = A sin(kx − ωt) = 12.0 cm · sin(1.57 rad/m · x − 31.4 rad/s · t)
(d) Differentiating with respect to time pulls out a chain-rule ω.
∂y
= Aω = 12.0 cm · 31.4 1/s = 3.77 m/s
∂t max
(e) Differentiating again with respect to time pulls out another ω.
2 ∂ y
= Aω 2 = 12.0 cm · (31.4 1/s)2 = 118 m/s2
∂t2 max
Problem 13.10. A transverse wave on a string is described by the wave function
π
y = (0.120 m) sin
x + 4πt .
8
(31)
(32)
(33)
(34)
(a) Determine the transverse speed and acceleration of an element of the string at t = 0.200 s for the point on the string
located at x = 1.60 m. (b) What are the wavelength, period, and speed of propagation of this wave?
(a) Notice that the phase φ = πx/8 + 4πt = 0.2π + 0.8π = π. Since the transverse displacement is a sin curve, that means
that the transverse velocity will be most negative and the acceleration will be zero.
∂y
(x = 1.60 m, t = 0.200 s) = Aω = −1.51 m/s
∂t
∂2y
(x = 1.60 m, t = 0.200 s) = 0 m/s2 .
∂t2
(35)
(36)
(b) To increase the phase by 2π, we need a ∆x2π = λ = 16.0 m. Similarly, we would need a ∆t2π = T = 0.500 s. The wave
speed then is v = ∆x/∆t = λ/T = 32.0 m/s.
Problem 13.11. A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction
with a speed of 30.0 m/s. At t = 0, an element of the string at x = 0 has a tranverse position of 2.00 cm and is traveling
downward with a speed of 2.00 m/s. (a) What is the amplitude of the wave? (b) What is the initial phase angle? (c) What
is the maximum transverse speed of an element of the string? (d) Write the wave function for the wave.
I think the easiest way to do this is backwards. (d) The transverse position and velocity must look something like
y = A cos[k(x + vt) + φ] = A cos(ωt + kx + φ)
∂y
= −ωA sin(ωt + kx + φ) ,
∂t
where the x + vt comes from the requirement for motion in the −x direction, and
part, we consider the situation at t = x = 0.
(b)
y(0, 0) = A cos(φ)
∂y
(0, 0) = −ωA sin(φ)
∂t
∂y
sin(φ)
∂t (0, 0)
= −ω
= −ω tan(φ)
y(0, 0)
cos(φ)
!
− ∂y
∂t (0, 0)
φ = arctan
.
ωy(0, 0)
∂y
∂t
(37)
(38)
comes from differentiating. In the next
(39)
(40)
(41)
(42)
From the period, we can find the angular velocity and wavenumber
ω=
2π
T
k=
ω
= 8.38 m/s
v
(43)
Plugging in for φ yields
φ = arctan
−(−2 m/s)
251 1/s · 0.0200 m
= arctan(0.398) = 0.379 rad .
(44)
Backside 180◦
Note that the tangent only has a range of [−π/2, pi/2], so you need to check and convince yourself that you aren’t on the
“backside” of the circle. We want a positive position and a negative velocity, which happens when both our sin and cos terms
are positive (see functions above), so we do want the first-quadrant answer for φ given above.
−
+
θ
−
+
(a) Now we can find the amplitude using trig
y(0, 0) = A cos(φ)
A=
(45)
y(0, 0)
= 2.15 cm .
cos(φ)
(46)
(c) And we can find the maximum transverse speed with
∂y
= Aω = 5.41 m/s .
∂t max
(47)
Problem 13.40. Two points A and B on the surface of the Earth are at the same longitude and 60.0◦ apart in latitude.
Suppose an earthquake at point A creates a P wave that reaches point B by traveling straight through the body of the Earth
at a constant speed of 7.80 km/s. The earthquake also radiates a Rayleigh wave, which travels along the surface of the Earth
at 4.50 km/s. (a) Which of these two seismic waves arrives at B first. (b) What is the time difference between the arrivals
of the two waves at B? Take the radius of the Earth to be 6370 km.
(a) The P wave reaches B first because it travels faster along a more direct route.
(b) To find the distances traveled by each wave, take a crossectional view of the Earth.
r
A
θ/2
B
Where the red line is the P wave path and the blue line is the Rayleigh wave path. From this picture it is clear that the
path lengths are
dP = 2r sin(θ/2)
dR = rθ
(48)
So the travel times are
dP
2 · 6370 km · sin(π/6)
=
= 817 s = 13.6 min
vP
7.80 km/s
dR
6370 km · π/3
tR =
=
= 1480 s = 24.7 min
vR
4.50 km/s
∆t = tR − tP = 665 s = 11.1 min
tP =
(49)
(50)
(51)
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