University Physics I - Classical Mechanics, 2019

11.2. SIMPLE HARMONIC MOTION 259
If we stick to using cosines, for definiteness, then the most general equation for the position of a
simple harmonic oscillator is as follows:
x(t) =A cos(ωt + φ) (11.10)
where φ is what we call a “phase angle,” that allows us to match the function to the initial
conditions—by which I mean, the object’s initial position and velocity. Specifically, you can see, by
setting t =0inEq.(11.10) and its derivative, that the initial position and velocity of the motion
described by Eq. (11.10) are
x i = A cos φ
v i = −ωA sin φ (11.11)
Conversely, if you are given x i and v i , you can use Eqs. (11.11) to determine A and φ, whichis
what you need to know in order to use Eq. (11.10) (note that the angular frequency, ω, doesnot
depend on the initial conditions—it is always the same regardless of how you choose to start the
motion). Specifically, you can verify that Eqs. (11.11) imply the following:
A 2 = x 2 i + v2 i
ω 2 (11.12)
and then, once you know A, you can get φ from either x i = A cos φ or v i = −ωA sin φ (in fact, since
the inverse sine and inverse cosine are both multivalued functions, you should use both equations,
to make sure you get the correct sign for φ).
11.2.1 Energy in simple harmonic motion
Equation (11.11) above actually follows from the conservation of energy principle for a harmonic
oscillator. Consider again the mass on the spring in Figure 11.3. Its kinetic energy is clearly
K = 1 2 mv2 , whereas the potential energy in the spring is 1 2 kx2 . Using Eq. (11.10) and its derivative,
we have
U spr = 1 2 kA2 cos 2 (ωt + φ)
K = 1 2 mω2 A 2 sin 2 (ωt + φ) (11.13)
Recalling Eq. (11.4), note that ω 2 = k/m, so if you substitute this in the second equation above,
you can see that the amplitude of both the potential and the kinetic energy is the same, namely,
1
2 kA2 . Since, for any angle θ, itisalwaystruethatcos 2 θ +sin 2 θ = 1, we find
E sys = U spr + K = 1 2 kA2 = 1 2 mω2 A 2 (11.14)