Homework 5 - Select Solutions
Harini Chandramouli
MATH2374
[email protected]
Section 4.1 - The Harmonic Oscillator
Problem 4. Find the simple harmonic motion described by the initial value problem.
ẍ + 4x = 0,
x(0) = 1,
ẋ(0) = −2.
Solution. This is an undamped (b = 0), unforced (f (t) = 0) oscillator. In this equation,
m = 1 and k = 4. So,
r
√
k
= 4 = 2.
ω0 =
m
Thus, the general solution is given by
x(t) = C1 cos(2t) + C2 sin(2t)
where C1 , C2 ∈ R. Taking the first derivative, we get
ẋ(t) = −2C1 sin(2t) + 2C2 cos(2t).
Applying our initial conditions, we get
x(0) = C1 = 1
ẋ(0) = 2C2 = −2.
Solvingg this system, we see that C1 = 1, C2 = −1. Thus the solution is
x(t) = cos(2t) − sin(2t).
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Problem 24. Determine the amplitude, phase angle, and period of the motion.
ẍ + x = 0,
x(0) = 1,
ẋ(0) = 1.
Solution. This is an undamped, unforced oscillator. Here, we have that m = 1 and k = 1.
Thus, ω0 = 1 rad/sec. The general solution is
x(t) = c1 cos(t) + c2 sin(t).
The natural frequency is then
f0 =
ω0
1
=
Hz
2π
2π
and the period is
T =
1
= 2π.
f0
Using the initial conditions, we get
x(0) = c1 = 1
ẋ(0) = c2 = 1.
Thus,
q
√
√
A = c21 + c22 = 12 + 12 = 2
π
tan(δ) = 1 =⇒ δ = .
4
√
Thus, we have that the amplitude A = 2, the phase angle δ = π4 , and the period of the
motion T = 2π.
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Problem 47. (Finding the Differential Equation) A mass of 500 gm is suspended from
the ceiling by a frictionless spring. The mass stretches the spring 50 cm in coming to its
equilibrium position, where the mass acting down is balanced exactly by the restoring force
acting up. The object is then pulled down an additional 10 cm and released.
(a) Formulate the initial value problem that describes the object’s motion, setting x equal
to the downward displacement from equilibrium.
(b) Solve for the motion of the object.
(c) Find the amplitude, phase angle, frequency, and period of the motion.
Solution.
(a) This problem is set up to use the CGS units (see Table 4.1.1 on pg. 198) but seeing as
those aren’t the ones I am used to using I will use the usual SI units. So first, let us
convert some things to the standard SI units:
m = 500g = 0.5kg
50cm = 0.5m
10cm = 0.1m.
Now, since x is equal tot he downward displacement, the downward direction is our
positive direction. Thus, the force acting on the mass is
m F = ma = (0.5kg) 9.8 2 = 4.9N.
sec
Now to find the spring constant k, we have to think about what it represents. It is the
strength of the spring, or force required to move the spring 1 unit of length, on our case
meters. Thus,
N
4.9N
= 9.8 .
k=
0.5m
m
Our initial conditions are given by x(0) = 0.1m and ẋ(0) = 1 assuming that our initial
velocity is 0. Thus, our initial value problem is
0.5ẍ + 9.8x = 0
x(0) = 0.1
ẋ(0) = 0
But we can make this look nicer, we can multiply through by 10 and we get the nicer
problem


5ẍ + 9.8x = 0
x(0) = 10


ẋ(0) = 0
(b) Here, since m = 5 and k = 98, we get
r
ω0 =
3
98
5
and so the solution is
r
x(t) = c1 cos
!
!
r
98
98
t + c2 sin
t
5
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where c1 , c2 ∈ R. We can solve for these constants using our initial conditions.
x(0) = c1 = 10
r
98
ẋ(0) =
c2 = 0 =⇒ c2 = 0.
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Thus,
r
x(t) = 10 cos
(c) The amplitude, A, is
A=
!
98
t .
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q
√
c21 + c22 = 102 + 02 = 10cm.
The phase angle, δ, is
tan(δ) =
c2
= 0 =⇒ δ = 0.
c1
The frequency, f0 , is
q
98
ω0
1
5
f0 =
=
=
2π
2π
2π
r
98
oscillations per sec.
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Lastly, the period of motion, T , is given to be
r
1
5
T =
sec.
= 2π
f0
98
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Note: These problems are taken from Differential Equations and Linear Algebra by Jerry
Farlow, James Hall, Jean McDill, and Beverly West. Specifically, I am following the Custom
Edition for the University of Minnesota Twin Cities.
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