Practice Problems for Midterm Exam - Mathematical Modeling | MATH 142, Exams of Mathematics

Material Type: Exam; Class: Mathematical Modeling; Subject: Mathematics; University: University of California - Los Angeles; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/31/2009

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Math 142 Practice problems for midterm exam
1. Phase plane sketch Consider a system with the potential V(x) = x2x4/4.
a. Find the stable and unstable equilibrium points.
Answer: Equilibrium points are where the derivative of the potential (and thus the force) is zero. This
occurs when 2xx3=0, which means x=0 and x=±2.
The equilibrium point is stable if the potential is concave up and unstable if it is concave down. With
V′′(x) = 23x2, we see that x=0 is stable and x=±2 is unstable.
b. Sketch the trajectories in the phase plane.
Answer:
c. Consider the potential V(x) = x2x4/4+2. Why does this potential produce the same dynamics?
Answer: The force is given by the derivative of the potential, which in both cases is f(x) = 2xx3. If the
force is the same, then Newton’s equation is the same, and so the dynamics are the same.
2. Damped harmonic oscillator If you were designing a suspension system for a car, would you want the
spring system to be underdamped, overdamped, or critically damped and why?
Answer: If the system is underdamped, then the car will continue to bounce after it hits a bump. If the
car is overdamped, the wheels will return to equilibrium slowly, so that two bumps in a row could cause
problems. Critically damped springs will return to equilibrium as fast as possible without oscillating.
3. Nonlinear pendulum Consider the non-linear pendulum ¨
θ=g
sinθ.
a. Compute the potential energy, using the above equation.
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Math 142 Practice problems for midterm exam

1. Phase plane sketch Consider a system with the potential V ( x ) = x^2 − x^4 /4. a. Find the stable and unstable equilibrium points. Answer: Equilibrium points are where the derivative of the potential (and thus the force) is zero. This occurs when 2 xx^3 = 0, which means x = 0 and x = ±

The equilibrium point is stable if the potential is concave up and unstable if it is concave down. With V ′′( x ) = 2 − 3 x^2 , we see that x = 0 is stable and x = ±

2 is unstable. b. Sketch the trajectories in the phase plane. Answer:

c. Consider the potential V ( x ) = x^2 − x^4 / 4 + 2. Why does this potential produce the same dynamics? Answer: The force is given by the derivative of the potential, which in both cases is f ( x ) = 2 xx^3. If the force is the same, then Newton’s equation is the same, and so the dynamics are the same.

2. Damped harmonic oscillator If you were designing a suspension system for a car, would you want the spring system to be underdamped, overdamped, or critically damped and why?

Answer: If the system is underdamped, then the car will continue to bounce after it hits a bump. If the car is overdamped, the wheels will return to equilibrium slowly, so that two bumps in a row could cause problems. Critically damped springs will return to equilibrium as fast as possible without oscillating.

3. Nonlinear pendulum Consider the non-linear pendulum ¨θ = − g ℓ sin θ. a. Compute the potential energy, using the above equation.

Answer:

V (θ) =

Z (^) θ

0

g ℓ sin θ d θ

= −

g ℓ cos θ

∣∣θ 0 = g

( 1 − cos θ)

It should be noted that this does not have the units of energy. To be a true energy, we need to multiply by m ℓ^2 , giving us

V (θ) = mg ℓ( 1 − cos θ)

b. The potential energy of a mass under the influence of gravity is given by V ( y ) = mg ( yy 0 ). Use this to compute then potential energy of the non-linear pendulum.

Answer:

L−Lcos( )θ

V (θ) = mg ℓ( 1 − cos θ)

c. Are they the same?

Answer: Yes.

d. Redo parts a and b for the non-dimensionalized non-linear pendulum.

Answer: Then non-dimensional non-linear pendulum is governed by ¨θ = − sin θ. So the only change is that the g /ℓ disappears. The potential is thus V (θ) = ( 1 − cos θ).

4. Exponential growth We saw in class two types of exponential growth: continuous and discrete. Suppose a population grows discretely with a yearly growth rate of R. If we modelled this with a continuous growth process, what instantaneous growth rate would produce the same doubling time?

Answer: The doubling time for discrete growth is given by

( 1 + Rt ) m^ = 2 m = ln 2 ln( 1 + R )

where we’ve let ∆ t = 1 yr. Continuous growth can be modeled as

N ( t ) = N 02

tt 0 td