Worksheet # 6 Problem 1. Avalanche forecasters measure the

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Worksheet # 6
Problem 1. Avalanche forecasters measure the temperature gradient
dT /dh, which is the rate at which the temperature in a snowpack T
changes with respect to its depth h. If the temperature gradient is large,
it may lead to a weak layer of snow in the snowpack. When these weak
layers collapse, avalanches occur. Avalanche forecasters use the following rule of thumb: If dT /dh exceeds 10 degrees C/meter anywhere in the
snowpack, conditions are favorable for weak layer formation and the risk
of avalanche increases.
Assume the temperature function is continuous and differentiable.
a. An avalanche forecaster digs a snow pit and takes two temperature
measurements. At the surface (h=0) the temperature is −12◦ C .
At a depth of 1.1 m the temperature is 2◦ C. Using the mean value
Theorem, what can he conclude about the temperature gradient? Is
the information of a weak layer likely?
b. One mile away, a skier finds that the temperature at a depth of 1.4m
is −1◦ C, and at the surface it is −12◦ C. What can be concluded
about the temperature gradient? Is the information of a weak layer
in her location likely?
c. Because snow is an excellent insulator, the temperature of snow covered ground is often 0◦ C. Furthermore, the surface temperature of
snow is a particular area does not vary much from one location to
another. Explain why a weak layer is more likely to form in places
where the snowpack is not too deep.
Problem 2. A flexible chain of length L is suspended between two
poles of equal height separated by a distance 2M (Figure below). By
x
Newtons laws, the chain describes a catenary y = a cosh( ) , where a is
a
the number such that L = 2a sinh( Ma ). The sag s is the vertical distance
from the highest to the lowest point on the chain; s = a cosh(M/a) − a.
Assume that M is fixed.
2
ds
da
da
(b) Calculate
using the relation L = 2a sinh( Ma ).
dL
ds
ds da
(c) Use (a) and (b) to compute
=
.
dL da dL
(d) Suppose that L = 160 meters and M = 50 meters. Use newton’s
method to find a value of a satisfying L = 2a sinh( Ma ).
(a) Calculate
(e) Use (c) and linear approximation to estimate the increase in the sag
∆s for changes in length ∆L = 1 and ∆L = 5.
Problem 3. The response of a circuit or other oscillatory system to an
input of frequency ω (omega) is described by the function
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φ(ω) = p 2
(ω0 − ω 2 )2 + 4D2 ω 2
Both ω0 (the natural frequency of the system) and D (the damping factor)
are positive constants. The graph of φ is called a resonance curve, and
the positive frequency ωr > 0, where φ takes its maximum
p value, if it
2
2
exists, is called
√ the resonant frequency. Show that ωr = ω0 − 2D is
0 < D < ω0 / 2 and that no resonant frequency exists otherwise.