15.4.1 Solving the Equation of Motion

The solution to an algebraic equation such as $x^{2} = 4$ is a number. The solution to a differential equation is a function. The x in Equation 15.32 is really $x (t)$ , the position as a function of time. The solution to this equation is a function $x (t)$ whose second derivative is the function itself multiplied by $(- k / m)$ .

One important property of differential equations that you will learn about in math is that the solutions are unique. That is, there is only one solution to Equation 15.32 that satisfies the initial conditions. If we were able to guess a solution, the uniqueness property would tell us that we had found the only solution. That might seem a rather strange way to solve equations, but in fact differential equations are frequently solved by using your knowledge of what the solution needs to look like to guess an appropriate function. Let us give it a try!

A photograph of L I G O Hanford data and predicted values of a gravitational wave in 2015. — The first observation of a gravitational wave in 2015 shows that space itself can oscillate.

We know from experimental evidence that the oscillatory motion of a spring appears to be sinusoidal. Let us guess that the solution to Equation 15.32 should have the functional form

x (t) = A cos (ω t + ϕ_{0})

(15.33)

where A, $ω$ , and $ϕ_{0}$ are unspecified constants that we can adjust to any values that might be necessary to satisfy the differential equation.

If you were to guess that a solution to the algebraic equation $x^{2} = 4$ is $x = 2$ , you would verify your guess by substituting it into the original equation to see if it works. We need to do the same thing here: Substitute our guess for $x (t)$ into Equation 15.32 to see if, for an appropriate choice of the three constants, it works. To do so, we need the second derivative of $x (t)$ . That is straightforward:

\begin{array}{l} x (t) = A cos (ω t + ϕ_{0}) \\ \frac{d x}{d t} = - ω A sin (ω t + ϕ_{0}) \\ \frac{d^{2} x}{d t^{2}} = - ω^{2} A cos (ω t + ϕ_{0}) \end{array}

(15.34)

If we now substitute the first and third of Equations 15.34 into Equation 15.32, we find

- ω^{2} A cos (ω t + ϕ_{0}) = - \frac{k}{m} A cos (ω t + ϕ_{0})

(15.35)

Equation 15.35 will be true at all instants of time if and only if $ω^{2} = k / m$ . There do not seem to be any restrictions on the two constants A and $ϕ_{0}$ —they are determined by the initial conditions.

So we have found—by guessing!—that the solution to the equation of motion for a mass oscillating on a spring is

x (t) = A cos (ω t + ϕ_{0})

(15.36)

where the angular frequency

ω = 2 π f = \sqrt{\frac{k}{m}}

(15.37)

is determined by the mass and the spring constant.

NOTE Once again we see that the oscillation frequency is independent of the amplitude A.

Equations 15.34 and 15.37 seem somewhat anticlimactic because we’ve been using these results for the last several pages. But keep in mind that we had been assuming $x = A cos ω t$ simply because the experimental observations “looked” like a cosine function. We’ve now justified that assumption by showing that Equation 15.36 really is the solution to Newton’s second law for a mass on a spring. The theory of oscillation, based on Hooke’s law for a spring and Newton’s second law, is in good agreement with the experimental observations.

EXAMPLE 15.5 Analyzing an oscillator

At $t = 0 s,$ a 500 g block oscillating on a spring is observed moving to the right at $x = 15 cm$ . It reaches a maximum displacement of 25 cm at $t = 0.30 s$ .

Draw a position-versus-time graph for one cycle of the motion.
What is the maximum force on the block, and what is the first time at which this occurs?

MODEL The motion is simple harmonic motion.

SOLVE

The position equation of the block is $x (t) = A cos (ω t + ϕ_{0})$ . We know that the amplitude is $A = 0.25 m$ and that $x_{0} = 0.15 m$ . From these two pieces of information we obtain the phase constant:

$ϕ_{0} = {cos}^{- 1} (\frac{x_{0}}{A}) = {cos}^{- 1} (0.60) = \pm 0.927 rad$

The object is initially moving to the right, which tells us that the phase constant must be between $- π$ and $0 rad$ . Thus $ϕ_{0} = - 0.927 rad$ . The block reaches its maximum displacement $x_{max} = A$ at time $t = 0.30 s$ . At that instant of time

$x_{max} = A = A cos (ω t + ϕ_{0})$

This equation can be true only if $cos (ω t + ϕ_{0}) = 1$ , which requires $ω t + ϕ_{0} = 0$ . Thus

$ω = \frac{- ϕ_{0}}{t} = \frac{- (- 0.927 rad)}{0.30 s} = 3.09 rad / s$

Now that we know $ω$ , it is straightforward to compute the period:

$T = \frac{2 π}{ω} = 2.0 s$

FIGURE 15.13 graphs $x (t) = (25 cm) cos (3.09 t - 0.927)$ , where t is in s, from $t = 0 s$ to $t = 2.0 s$ .

FIGURE 15.13 Position-versus-time graph for the oscillator of Example 15.5.

FIGURE 15.13 Full Alternative Text
We found the acceleration of SHM to be $a_{x} = - ω^{2} x$ , so the force on the block at position x is $F_{x} = - m ω^{2} x$ . The force will be a maximum, $F_{max} = m ω^{2} A$ , when x reaches its minimum displacement $x = - A$ . For the block,

$F_{max} = (0.50 kg)(3.09 {rad/s)}^{2} (0.25 m) = 1.2 N$

This occurs exactly half a period (1.0 s) after the block reaches its maximum displacement, thus at $t = 1.3 s$ .

REVIEW A 2 s period is a modest oscillation, so we don’t expect the block’s acceleration to be extreme. A maximum force of 1.2 N on a 0.5 kg block causes a maximum acceleration of $2.4 {m/s}^{2}$ , which seems reasonable.