# 8.6: The general Solution to the Equation

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The general Solution to the Equation A more general way to write this equation that emphasizes its applicability to all cases of SHM is as follows: We will arbitrarily use the symbol “y”, but we just as well have used \( \theta \), or x, or any other symbol.

\[ \dfrac{d^2}{dt^2} y(t) = \left(\dfrac{- 2 \pi} { T} \right)^2 y(t) \]

The general solution in standard form is,

\[ y(t) = A \sin \left( \dfrac{2 \pi t}{T} + \phi \right) \]

What is the meaning of the constants? The sine function goes through a complete cycle every \(2\pi\) radians. This occurs each time *t * increases by an amount * T* . Thus

*is the*

**T***period*of the oscillations, as was mentioned before. The reciprocal of the period is the

*of oscillations, \( \mathcal{f} \):*

**frequency**\[ \mathcal{f} = \dfrac{1}{T}\]

* T* is the time required to complete the cycle, while \(\mathcal{f}\) is the number of cycles per second.

*is measured in seconds;\(\mathcal{f}\), in reciprocal seconds (1/s), which are called hertz and abbreviated Hz.*

**T**The maximum value of the sine function is +1 and the minimum value is -1. Thus, A is the * amplitude* of the oscillations. That is, the maximum value of y is +A and the minimum value is -A .

The angle \( \phi \) is determined by the value of y at the particular time t = 0. If y has its positive maximum value at t = 0, then \(\phi \) has to be \( 90^\circ\) or \(\pi/2 \) radians. We say that \(\phi\) depends on the “initial conditions”. The angle \(\phi\) is often n referred to as the ** phase angle**. By including the phase angle, we can make the sine function fit any particular physical situation. Without the phase angle, we would always have to start timing the oscillation when the position had the value zero. By including the phase angle, we have a perfectly general solution.

The solutions we have written down describe the position as a function of time for any object vibrating in simple harmonic motion. They give the specific time dependence of the position of whatever it is that is vibrating. The three constants depend on the particular situation.

Let's explore our solution to SHM further. The *equillibrium *value of y is the value y has when no oscillation is occurring. For the way we have written the solutions, this value is zero. Thus, the amplitude is the change in y that occurs in going from the equilibrium value to the maximum value of y.* *

The picture(Figure 8.6.1) is for the angle \(\phi = 0 \). Changing the value of the phase angle \(\phi \) shifts the curve sideways. Compare this plot to the plot of the sine solution with \(\phi = \pi/4 \) radians, in the following graph. You can see that they are just the same except for a sideways shift.

**Figure 8.6.1**

To summarize, then, the solutions we have written down describe the position as a function of time for any object vibrating in simple harmonic motion. They give the specific time dependence of the position of whatever it is that is vibrating. The three constants, ** A**,

**and \(\phi\) characterize the motion and depend on the particular situation.**

*T***Figure 8.6.2**

The period, * T*, or frequency, \( \mathcal{f} \), of an oscillating system is determined by the constants appearing as the coefficient of the linear term of the differential equation when written in standard form:

\[ \dfrac{d^2}{dt^2} y(t) = \left(\dfrac{- 2 \pi} { T} \right)^2 y(t) \]

For a mass on a spring, we found that,

\[ (\dfrac{2\pi}{T})^2 = \dfrac{k}{m} \rightarrow T = 2\pi \sqrt{\dfrac{m}{k}} \]

For the pendulum,

\[ (\dfrac{2\pi}{T})^2 = \dfrac{g}{l} \rightarrow T = 2\pi \sqrt{\dfrac{g}{l}} \]

Note that you do not need to write out the solution of the differential equation to get the frequency or period. It comes directly from applying Newton’s 2^{nd} law and “reading off” the constants.

### Energy for a Mass Hanging from a Spring

As a mass hanging from a spring oscillates, very little mechanical energy is converted to thermal energy or sound, so we expect the mechanical energy to remain essentially constant for many periods.

The potential energy is defined in terms of the work required to move the spring, and has the value,

\[ PE = \dfrac{1}{2} ky^2 \]

assuming that the origin is placed at the equilibrium value of y and that the potential energy is defined as zero there.

As usual kinetic energy is,

\[ KE = \dfrac{1}{2} mv^2 \]

Let's substitute the general solution for \(y(t)\) into these expressions:

\[ PE = \dfrac{1}{2}ky^2 = \dfrac{1}{2}kA^2 \sin \left( \dfrac{2 \pi t}{T} + \phi \right) \]

To calculate the kinetic energy, we differentiate the expression for y(t) to get v(t):

\[ v(t) = \dfrac{dy}{dt} = \dfrac{2\pi}{T}A \cos \left( \dfrac{2 \pi t}{T} + \phi \right) \]

so that

\[ KE = \dfrac{1}{2}m \left(\dfrac{ 2\pi}{T} \right)^2A^2 \cos^2 \left( \dfrac{2 \pi t}{T} + \phi \right) \]

But for a mass spring system, (2 \(\pi\)/T)^{2} = k/m. We can substitute this relation into the equation 8.6.11, finally getting the expression for the kinetic energy:

\[ KE = \dfrac{1}{2}k A^2 \cos^2 \left( \dfrac{2 \pi t}{T} + \phi \right) \]

We note that the maximum values of the PE and KE are the same. Also, the time average values of the KE and the PE are the same: 1/2 of the maximum values, since the average of \(sin^2\) or \(cos^2\) is just 1/2. In Chapter 3 we could merely say this was plausible in our discussion of equipartition of energy. Now we see why we were justified in saying the average energy in the KE mode is the same as the average in the PE mode.

When we add the kinetic energy to the potential energy to get the total energy, we notice that we have the sum \((sin^2 + cos^2) \) which always has value unity. Hence the total energy is,

\[E_{tot}= 1/2A^2 \]

There are two things that are important about this equation 8.6.13,

- the total energy is a constant just as we expected, and
- the total energy is proportional to the square of the amplitude.

This is a characteristic of all kinds of oscillating systems and extends even to wave motions, as will see in the next chapter in Part 3 of this text.

### Energy Graphs

The Potential Energy Function,

\[PE = \frac{1}{2}y^2\]

is a parabola when plotted as a function of y, while the total energy, being constant, is a horizontal line (Figure 8.4.3).

**Figure 8.6.3**

The kinetic energy is the difference, \(KE=E_{tot} - PE\). But the kinetic energy cannot be negative, since the mass is never negative and \(v^2\) is never negative regardless of the sign of \(v\) itself. This means that the oscillation is limited, and can go from \(y_{max}\) to \(y_{min} \). Of course this just reinforces what we already know, since \(y_{max} = A\) and \(y_{min}=-A\).

But for any object that oscillates about an equilibrium position, even when the force law is not as simple as F=-ky, this graphical analysis provides an accurate description for small oscillations. As an example, look at the plot below(Figure 8.6.4):

**Figure 8.6.4**

This is the shape of the potential energy between two bound atoms that we encountered in chapter 3. Even though the pair-wise potential energy function is not a parabola over all distances, near the minimum it is approximately so, and therefore the oscillations of this system will be *simple harmonic *if they are at sufficiently small amplitude about the minimum in the potential energy curve.

Thus, we can make a very strong statement: essentially every system that vibrates, does so in SHM for small amplitudes of vibration. All molecules, including those in our bodies, and all atoms in solids and liquids move essentially in simple harmonic motion.

### Applying our Results to the Universe!

Let’s now consider our model for matter. For liquids and solids, we picture molecules that are bound to each other as if they have little springs attaching them together. They bounce around, and in liquids tumble and change positions. What is the nature of the oscillations these molecules undergo? And what happens when we exert external forces on the matter and bend it or compress it. How then does the matter respond on a macroscopic scale? What we saw in the discussion above on SHM, is that if the restoring force is proportional to the displacement, but in the opposite direction of the displacement, then SHM results. portional to the displacement, but in the opposite direction of the displacement, then SHM results. This will always be the case if the displacement from equilibrium is sufficiently small. The result is that everything, on both the atomic scale and macroscopic scale, tends to vibrate in SHM. This is very nice, because we know the solution! The period of oscillation depends on factors like the mass of the particles or object being considered and the strength of the restoring forces. If we can identify the forces acting and write down Newton’s 2^{nd} law, then for small oscillations, we have the problem solved!

Think about what we have accomplished. For any kind of matter, we know how to go about finding its vibration frequencies when subjected to external forces as well as how its internal parts oscillate. We have a very general and powerful approach that works for almost all vibrating phenomena on any scale!