22. Applications of Second order Differential Equations

Dated: 02-12-2024

A single differential equation can serve as mathematical model for many different phenomena in science and engineering.
Different forms of the 2nd order linear differential equation

\[a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x)\]

appear in the analysis of problems in physics, chemistry and biology.

Simple Harmonic Motion

When the Newton’s 2nd law is combined with the Hook’s Law, we can derive a differential equation governing the motion of a mass attached to spring–the simple harmonic motion.

Hook's Law

Suppose that

• A mass is attached to a flexible spring suspended from a rigid support.
• The spring stretches by an amount \(s\).
• The spring exerts a restoring \(F\) opposite to the direction of elongation or stretch.

The hook's law states that the force \(F\) is proportional to the elongation \(s\)

\[F = ks\]

Where \(k\) is called the spring constant.

Note that

• Different masses stretch a string by different amount i.e. \(s\) is different for different \(m\).
• The spring is characterized by the spring constant \(k\).
• For example if \(W = 10 \text{ lbs}\) and \(s = \frac 1 2 \text{ ft}\)
  Then

\[F = ks\]

\[10 = \frac 1 2 k\]

\[k = 20 \frac {\text{lbs}}{\text{ft}}\]

If \(W = 8 \text{ lbs}\) then

\[8 = 20(s) \implies s = \frac 2 5 \text{ ft}\]

Newton's Second Law

When a force \(F\) acts upon a body, the acceleration \(a\) is produced in the direction of the force whose magnitude is proportional to the magnitude of force

\[F = ma\]

Where \(m\) is constant of proportionality and it represents mass of the body.

Weight

The gravitational force exerted by the earth on a body of mass \(m\) is called weight of the body, denoted by \(W\).
In the absence of air resistance, the only force acting on a freely falling body is its weight. Thus from Newton’s 2nd law of motion

\[W = mg\]

Where \(m\) is measured in slugs, kilograms or grams and

\[g = 32 \frac {ft}{s^2}, 9.8 \frac m {s^2}, 980 \frac {cm}{s^2}\]

Differential Equation

When a body of mass \(m\) is attached to a spring
The spring stretches by an amount \(s\) and attains an equilibrium position.
At the equilibrium position, the weight is balanced by the restoring force \(ks\).
Thus, the condition of equilibrium is

\[mg = ks \implies mg - ks = 0\]

If the mass is displaced by an amount \(x\) from its equilibrium position and then released.
The restoring force becomes \(k(s + x)\).
So the resultant of weight and the restoring force acting on the body is given by

\[\text{Resultant } = -k(s + x) + mg\]

By Newton’s 2nd Law of motion, we can written

\[m\frac{d^2x}{dt^2}=-k(s+x)+mg\]

\[m\frac{d^2x}{dt^2}=-kx-ks+mg\]

\[mg-ks=0\]

\[\therefore m\frac{d^2x}{dt^2}=-kx\]

The negative indicates that the restoring force of the spring acts opposite to the direction of motion.
The displacements measured below the equilibrium position are positive.
By dividing with \(m\) , the last equation can be written as

\[\frac{d^2x}{dt^2}+\frac{k}{m}x=0\]

\[\frac{d^2x}{dt^2}+\omega^2x=0\]

Where \(\omega^2 = \frac k m\).
This equation is known as the equation of simple harmonic motion or as the free un-damped motion.

Initial Conditions

Associated with the differential equation

\[\frac{d^2x}{dt^2}+\omega^2x=0\]

are the obvious initial conditions

\[x(0)=\alpha, \quad x'(0)=\beta\]

These initial conditions represent the initial displacement and the initial velocity. For example

• If \(\alpha > 0, \beta < 0\) then the body starts from a point below the equilibrium position with an imparted upward velocity.
• If \(\alpha < 0, \beta = 0\) then the body starts from rest \(|\alpha|\) units above the equilibrium position.

Solution and Equation of Motion

Consider the equation of simple harmonic motion

\[\frac{d^2x}{dt^2}+\omega^2x=0\]

Put

\[x=e^{mx}, \quad \frac{d^2x}{dt^2}=m^2e^{mx}\]

Then the auxiliary equation is

\[m^2+\omega^2=0\implies m=\pm \iota \cdot \omega\]

Thus the auxiliary equation has complex roots.

\[m_1=\omega i, \quad m_2=-\omega i\]

Hence, the general solution of the equation of simple harmonic motion is

\[x(t)=c_1\cos\omega t+c_2\sin\omega t\]

Alternate Form of Solution

It is often convenient to write the above solution in a alternative simpler form.
Consider

\[x(t)=c_1\cos\omega t+c_2\sin\omega t\]

and suppose that \(A, \phi \in \mathbb R\) such that

\[c_1=A\sin\phi, \quad c_2=A\cos\phi\]

\[A=\sqrt{c_1^2+c_2^2}, \quad \tan\phi=\frac{c_1}{c_2}\]

\[x(t)=A \sin (\omega t) \cos(\phi) + B\cos(\omega t)\sin(\phi)\]

\[x(t)=A\sin(\omega t+\phi)\]

The number \(\phi\) is called the phase angle.

Note that

This form of the solution of the equation of simple harmonic motion is very useful because

• Amplitude of free vibrations becomes very obvious
• The times when the body crosses equilibrium position are given by

\[x=0\Rightarrow \sin(\omega t+\phi)=0\]

\[\omega t+\phi=n\pi\]

Where \(n\) is a non negative integer.

The Nature of Harmonic Motion

Amplitude

We know that the solution of the equation of simple harmonic motion can be written as

\[x(t)=A\sin(\omega t+\phi)\]

Clearly, the maximum distance that the suspended body can travel on either side of the equilibrium position is \(A\).
This maximum distance called the amplitude of motion and is given by

\[\text{Amplitude } = A = \sqrt{c_1^2+c_2^2}\]

A Vibration or a Cycle

In travelling from \(x = A\) to \(x = -A\) and then back to \(A\), the vibrating body completes one vibration or one cycle.

Period of Vibration

The simple harmonic motion of the suspended body is periodic and it repeats its position after a specific time period \(T\).
We know that the distance of the mass at any time \(t\) is given by

\[x=A\sin(\omega t+\phi)\]

\[\because A\sin\left[\omega\left(t+\frac{2\pi}{\omega}\right)+\phi\right]\]

\[=A\sin\left[(\omega t+\phi+2\pi)\right]\]

\[=A\sin[(\omega t+\phi)]\]

Therefore, the distances of the suspended body from the equilibrium position at the times \(t\) and \(t + \frac {2 \pi} \omega\) are same.
Further, velocity of the body at any time \(t\) is given by

\[\frac{dx}{dt}=A\omega\cos(\omega t+\phi)\]

\[A\omega\cos\left[\omega\left(t+\frac{2\pi}{\omega}\right)+\phi\right]\]

\[=A\omega\cos[\omega t+\phi+2\pi]\]

\[=A\omega\cos(\omega t+\phi)\]

Therefore the velocity of the body remains unaltered if \(t\) is increased by \(\frac {2 \pi} \omega\).
Hence the time period of free vibrations described by the 2nd order differential equation

\[\frac{d^2x}{dt^2}+\omega^2x=0\]

is given by

\[T = \frac {2 \pi}{\omega}\]

Frequency

The number of vibrations or cycles completed in unit time is known as frequency of the free vibrations, denoted by \(f\).
Since the cycles completed in time \(T\) is \(1\).
Therefore, the number of cycles completed in a unit of time is \(\frac 1 T\).

\[f=\frac{1}{T}=\frac{\omega}{2\pi}\]

Example

Solve and interpret the initial value problem

\[\frac{d^2x}{dt^2}+16x=0\]

\[x(0)=10, \quad x'(0)=0\]

Interpretation

Comparing the initial conditions

\[x(0)=10, \quad x'(0)=0\]

with

\[x(0)=\alpha, \quad x'(0)=\beta\]

we see that

\[\alpha=10, \quad \beta=0\]

Thus the problem is equivalent to

• Pulling the mass on a spring 10 units below the equilibrium position.
• Holding it there until time t = 0 and then releasing the mass from rest.

Solution

Consider the differential equation

\[\frac{d^2x}{dt^2}+16x=0\]

Put

\[x=e^{mt}, \quad \frac{d^2x}{dt^2}=m^2e^{mt}\]

Then, the auxiliary equation is

\[m^2+16=0\]

\[\implies m=0\pm4i\]

Therefore, the general solution is:

\[x(t)=c_1\cos4t+c_2\sin4t\]

Now we apply the initial conditions.

\[x(0)=10\implies c_1.1+c_2.0=10\]

\[\therefore c_1=10\]

So that

\[x(t)=10\cos4t+c_2\sin4t\]

\[\frac{dx}{dt}=-40\sin4t+4c_2\cos4t\]

\[\therefore x'(0)=0\implies -40(0)+4c_2.1=0\]

\[\therefore c_2=0\]

Hence, the solution of the initial value problem is

\[x(t) = 10 \cos 4t\]

Note that

• Clearly, the solution shows that once the system is set into motion, it stays in motion with mass bouncing back and forth with amplitude being 10 units.
• Since \(\omega = 4\), therefore, the period of oscillations is

\[T=\frac{2\pi}{4}=\frac{\pi}{2}\text{ seconds}\]

References

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