There are many applications of 2nd order differential equations, particularly 2nd order linear. We take a look at a couple.

Springs: Consider an object of constant mass \(m>0\) suspended from a spring (of negligible mass). We say that the spring-mass system is at equilibrium when the object is at rest and the sum of the forces acting on it is \(0\). The corresponding position is called the equilibrium position.

Let \(y\) be the displacement of the mass from the equilibrium where \(y>0\) is upward.

The following forces act on the object:

• Gravity: \(F_{g}=-mg\) where \(g\) is the gravitational constant.
  
• Spring Force (Hooke’s Law): \(F_{s}=k\Delta L\) where \(k>0\) is the spring constant – sometimes called the \(``\)spring stiffness" – and \(\Delta L\) is the change in length (stretching is positive) from the spring equilibrium position with no mass attached. This is often called the \(``\)natural length of the spring."
  

The following forces might act on the object depending on the situation:

• Damping Force: \(F_{d}=-cy'\) where \(c>0\) is the damping constant (due to medium resistance, friction, or even a mechanical damper).
  
• External force(s): A time-dependent force independent of the motion, which we denote by \(F(t)\), could act on the object.
  

By Newton’s second law we can write a de to model the spring-mass system described above:

Let’s relate \(\Delta L\) and \(y\). Imagine the spring without a mass attached. When a mass \(m\) is attached, the spring stretches by some amount \(\ell \) from its natural length to the spring-mass equilibrium position, where the sum of forces on the spring is \(0\), so \(F_{g}+F_{s}=0\). Hence \(-mg+k\ell =0\), or \(\ell =mg/k\).

If at this point, a displacement of \(y\) occurs then we would have \(\Delta L=\ell -y=mg/k-y\).

So \(my''=F(t)-mg-cy'+k\Delta L\) becomes \(my''=F(t)-cy'-ky\) or equivalently \(my''+cy'+ky=F(t)\).

If \(F(t)=0\) we say the motion is free. Otherwise, it is said to be forced.

If \(c=0\) we say the motion is undamped. If \(c>0\) we say the motion is damped. Simple harmonic motion or a simple harmonic oscillator (SHO) is free and undamped.

We begin with the case of a spring-mass system that is a simple harmonic oscillator:

Consider the free undamped system modeled by \(my''+ky=0\). The characteristic equation for this constant-coefficient homogeneous de is \(mr^2+k=0\) and has roots at \(r=\pm \sqrt {\frac {k}{m}}\,i\). It is standard notation to let \(\omega =\sqrt {\frac {k}{m}}\). So we know that a general solution is

Let \(A=\sqrt {c_1^2+c_2^2}\). If \(A=0\) then \(y=0\).

Otherwise, we can rewrite the above as \(y=A\left (\frac {c_{1}}{A}\cos {(\omega t)}+\frac {c_{2}}{A}\sin {(\omega t)}\right )\). There exists
\(\varphi \in [0,2\pi )\) satisfying \(c_1=A\cos {(\varphi )}\) and \(c_2=A\sin {(\varphi )}\).

So we can rewrite the above as \(y=A\left (\cos {(\varphi )}\cos {(\omega t)}+\sin {(\varphi )}\sin {(\omega t)}\right )\). Using the Trigonometric identity \(\cos {(\alpha -\beta )}=\cos {(\alpha )}\cos {(\beta )}+\sin {(\alpha )}\sin {(\beta )}\), where \(\alpha =\omega t\) and \(\beta =\varphi \) we get \(y=A\cos {(\omega t-\varphi )}\) which even works if \(A=0\). Note: \(A\cos {(\omega t-\varphi )}=A\sin {\left (\omega t-\left [\varphi -\frac {\pi }{2}\right ]\right )}\).

Here are some key features of a simple harmonic oscillator where \(y=A\cos {(\omega t-\varphi )}\):

\(A\) is the amplitude of the oscillations – which range between \(-A\) and \(A\). \(\omega \) is the angular frequency – in radians per second. \(f=\frac {\omega }{2\pi }\) is the frequency – in cycles per second, or hertz (hz). \(T=\frac {1}{f}=\frac {2\pi }{\omega }\) is the period – in seconds. \(\varphi \) is the phase angle – in radians. \(c=\frac {\varphi }{\omega }\) is the phase shift – in seconds.

Example: Suppose an object is hung from a spring whose natural length is \(98\) cm and attains an equilibrium position of \(1\) m. Suppose the object is displaced downward \(5\) cm, but given an initial upward velocity of \(11\) cm/s. Find the displacement for time \(t>0\), and then the amplitude, frequency and period of the oscillations.

Since \(g/\ell =k/m\), we have that \(\omega =\sqrt {k/m}=\sqrt {g/\ell }=\sqrt {9.8/0.02}\approx 22\). So
\(y\approx c_{1}\cos {(22t)}+c_{2}\sin {(22t)}\). Imposing \(y(0)=-0.05\) yields \(c_{1}=-0.05\). Imposing \(y'(0)\approx 0.11\) yields \(22c_{2}\approx 0.11\), so \(c_{2}\approx 0.005\).

Hence \(y\approx -0.05\cos {(22t)}+0.005\sin {(22t)}\).

The amplitude is \(A\approx \sqrt {(-0.05)^2+(0.005)^2}\approx 0.05\) m, the frequency is \(f=\frac {\omega }{2\pi }\approx \frac {11}{\pi }\) hz & the period is \(T=\frac {1}{f}\approx \frac {\pi }{11}\) sec (the phase angle would be \(\varphi \approx \tan ^{-1}{(-0.1)}+\pi \approx 3.04\) rad and the phase shift would be approximately 0.28 s).

Let’s consider an example with an external force: A mass of \(m\) kg is hung on a large spring, of spring constant \(k\) N/m, whose natural length is \(50\) cm and is brought to rest at an equilibrium position of \(77.2\) cm. Starting at rest at time \(t=0\), Someone pushes the mass with a force of \(F(t)=6m\cos {(6t)}\) N at time \(t\geq 0\) seconds. Let’s find the displacement \(y(t)\) above equilibrium at time \(t\geq 0\) seconds and describe the long-run behavior.

Since \(g/\ell =k/m\), we have that \(\omega =\sqrt {k/m}=\sqrt {g/\ell }=\sqrt {9.8/0.272}\approx 6\).

We can model this motion by \(my''+ky=F(t)\), \(y(0)=0\), \(y'(0)=0\). So \(my''+ky=6m \cos {(6t)}\) or equivalently \(y''+36 y= 6\cos {(6t)}\). The functions \(y_1=\cos {(6t)}\) and \(y_2=\sin {(6t)}\) form a fundamental set of solutions to the complementary equation. By using Undetermined Coefficients, Reduction of Order, of Variation of Parameters, we get that \(y_p=0.5t\sin {(6t)}\) is a solution. The general solution is \(y=c_1 \cos {(6t)}+(0.5t +c_2)\sin {(6t)}\). By using the initial conditions, \(c_1=0\) and \(c_2=0\). So \(y=0.5t \sin {(6t)}\). In the long run, the oscillations get larger and larger (until the spring eventually fails). This is because the mass is forced with a periodic force that is in sync with the natural frequency of the spring mass system. This concept is called \(``\)resonance."

Now let’s consider a free damped system: \(my''+cy'+ky=0\) where \(c>0\). The characteristic polynomial is \(mr^{2}+cr+k\). The roots of this polynomial are \(\displaystyle r=\frac {-c\pm \sqrt {c^{2}-4mk}}{2m}\).

Underdamped case: \(c<\sqrt {4mk}\)

Then we get two complex conjugate roots \(\alpha \pm \omega i\) where \(\alpha =-c/(2m)\) and \(\omega =\dfrac {\sqrt {4mk-c^{2}}}{2m}\).

Hence a general solution is \(y=e^{\alpha t}(c_{1}\cos {(\omega t)}+c_{2}\sin {(\omega t)})\) and since \(\alpha <0\) we get that \(y\rightarrow 0\) as \(t\rightarrow \infty \), but with oscillations.

Overdamped case: \(c>\sqrt {4mk}\)

Then we get two distinct real and negative roots \(r_{1},r_{2}\), and a general solution of
\(y=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}\) which goes towards \(0\) as \(t\rightarrow \infty \). In this case no oscillation occurs as \(y\neq 0\) cannot equal zero for more than one value of \(t\).

Critically damped case: \(c=\sqrt {4mk}\)

Then we get one repeated real and negative root \(r=-c/(2m)\), and a general solution of
\(y=(c_{1}+c_{2}t)e^{rt}\) which goes towards \(0\) as \(t\rightarrow \infty \). As in the overdamped case, no oscillation occurs and \(y\neq 0\) cannot equal zero for more than one value of \(t\).