15.2.1 Initial Conditions: The Phase Constant

Now we’re ready to consider the possibility of other initial conditions. The particle in Figure 15.5 started at $ϕ_{0} = 0$ . This was equivalent to an oscillator starting at the far right edge, $x_{0} = A$ . FIGURE 15.6 shows a more general situation in which the initial angle $ϕ_{0}$ can have any value. The angle at a later time t is then

ϕ = ω t + ϕ_{0}

(15.13)

In this case, the particle’s projection onto the x-axis at time t is

x (t) = A cos (ω t + ϕ_{0})

(15.14)

The figure illustrates the before and after positions of a particle in a uniform circular motion on an x y coordinate plane.

FIGURE 15.6 Full Alternative Text

If Equation 15.14 describes the particle’s projection, then it must also be the position of an oscillator in simple harmonic motion. The oscillator’s velocity $v_{x}$ is found by taking the derivative $d x / d t$ . The resulting equations,

\begin{array}{l} x (t) = A cos (ω t + ϕ_{0}) \\ v_{x} (t) = − ω A sin (ω t + ϕ_{0}) = − v_{\max} sin (ω t + ϕ_{0}) \end{array}

(15.15)

are the two primary kinematic equations of simple harmonic motion.

The quantity $ϕ = ω t + ϕ_{0}$ , which steadily increases with time, is called the phase of the oscillation. The phase is simply the angle of the circular-motion particle whose shadow matches the oscillator. The constant $ϕ_{0}$ is called the phase constant. It is determined by the initial conditions of the oscillator.

To see what the phase constant means, set $t = 0$ in Equations 15.15:

\begin{matrix} x_{0} = A cos ϕ_{0} \\ v_{0 x} = - ω A \sin ϕ_{0} \end{matrix}

(15.16)

The position $x_{0}$ and velocity $v_{0 x}$ at $t = 0$ are the initial conditions. Different values of the phase constant correspond to different starting points on the circle and thus to different initial conditions.

The cosine function of Figure 15.3 and the equation $x (t) = A cos ω t$ are for an oscillation with $ϕ_{0} = 0 rad$ . You can see from Equations 15.16 that $ϕ_{0} = 0 rad$ implies $x_{0} = A$ and $v_{0} = 0$ . That is, the particle starts from rest at the point of maximum displacement.

FIGURE 15.7, , illustrates these ideas by looking at three values of the phase constant: $ϕ_{0} = π / 3 rad$ $(60 °)$ , $- π / 3 rad$ $(- 60 °)$ , and $π rad$ $(180 °)$ . Notice that $ϕ_{0} = π / 3 rad$ and $ϕ_{0} = - π / 3 rad$ have the same starting position, $x_{0} = \frac{1}{2} A$ . This is a property of the cosine function in Equations 15.16. But these are not the same initial conditions. In one case the oscillator starts at $\frac{1}{2} A$ while moving to the left, in the other case it starts at $\frac{1}{2} A$ while moving to the right. You can distinguish between the two by visualizing the motion.

The figure consists of three parts that illustrate a particle in a uniform circular motion, block in simple harmonic motion, and position and velocity versus time graphs for the motion with different phase constants.

FIGURE 15.7 Full Alternative Text

All values of the phase constant $ϕ_{0}$ between 0 and $π rad$ correspond to a particle in the upper half of the circle and moving to the left. Thus $v_{0 x}$ is negative. All values of the phase constant $ϕ_{0}$ between $π$ and $2 π rad$ (or, as they are usually stated, between $- π$ and 0 rad) have the particle in the lower half of the circle and moving to the right. Thus $v_{0 x}$ is positive. If you’re told that the oscillator is at $x = \frac{1}{2} A$ and moving to the right at $t = 0$ , then the phase constant must be $ϕ_{0} = - π / 3 rad,$ not $+ π / 3 rad$ .

EXAMPLE 15.3 Using the initial conditions

An object on a spring oscillates with a period of 0.80 s and an amplitude of 10 cm. At $t = 0 s,$ it is 5.0 cm to the left of equilibrium and moving to the left. What are its position and direction of motion at $t = 2.0 s ?$

MODEL An object oscillating on a spring is in simple harmonic motion.

SOLVE We can find the phase constant $ϕ_{0}$ from the initial condition $x_{0} = - 5.0 cm = A cos ϕ_{0}$ . This condition gives

ϕ_{0} = {cos}^{- 1} (\frac{x_{0}}{A}) = {cos}^{- 1} (- \frac{1}{2}) = \pm \frac{2}{3} π rad = \pm 120 °

Because the oscillator is moving to the left at $t = 0$ , it is in the upper half of the circular-motion diagram and must have a phase constant between 0 and $π rad$ . Thus $ϕ_{0}$ is $\frac{2}{3} π rad$ . The angular frequency is

ω = \frac{2 π}{T} = \frac{2 π}{0.80 s} = 7.85 rad / s

Thus the object’s position at time $t = 2.0 s$ is

\begin{matrix} x (t) & = & A cos (ω t + ϕ_{0}) \\ = & (10 cm) cos ((7.85 rad / s) (2.0 s) + \frac{2}{3} π) \\ = & (10 cm) cos (17.8 rad) = 5.0 cm \end{matrix}

The object is now 5.0 cm to the right of equilibrium. But which way is it moving? There are two ways to find out. The direct way is to calculate the velocity at $t = 2.0 s :$

v_{x} = - ω A sin (ω t + ϕ_{0}) = + 68 cm / s

The velocity is positive, so the motion is to the right. Alternatively, we could note that the phase at $t = 2.0 s$ is $ϕ = 17.8 rad$ . Dividing by $π$ , you can see that

ϕ = 17.8 rad = 5.67 π rad = (4 π + 1.67 π) rad

The $4 π$ rad represents two complete revolutions. The “extra” phase of $1.67 π rad$ falls between $π$ and $2 π rad,$ so the particle in the circular-motion diagram is in the lower half of the circle and moving to the right.

NOTE The inverse-cosine function ${cos}^{- 1}$ is a two-valued function. Your calculator returns a single value, an angle between 0 rad and $π rad$ . But the negative of this angle is also a solution. As Example 15.3 demonstrates, you must use additional information to choose between them.

15.2.1 Initial Conditions: The Phase Constant

FIGURE 15.6 A particle in uniform circular motion with initial angle $ϕ_{0}$ .

FIGURE 15.7 Different initial conditions are described by different values of the phase constant.

EXAMPLE 15.3 Using the initial conditions