# 5.5: Problems

Problem $$\PageIndex{1}$$

Solve the following initial value problems:

1. $$\frac{d^2x}{dt^2}+\frac{dx}{dt}-2x=0; \;x(0)=1; \;x'(0)=0$$
2. $$\frac{d^2x}{dt^2}+6\frac{dx}{dt}+9x=0; \;x(1)=0; \;x'(1)=1$$
3. $$\frac{d^2x}{dt^2}+9x=0; \;x(\pi/3)=0; \;x'(\pi/3)=-1$$
4. $$\frac{d^2x}{dt^2}-2\frac{dx}{dt}+2x=0; \;x(0)=1; \;x'(0)=0$$

Problem $$\PageIndex{2}$$

The simple harmonic oscillator consists of a body moving in a straight line under the influence of a force whose magnitude is proportional to the displacement $$x$$ of the body from the point of equilibrium, and whose direction is towards this point. $\label{ode2:spring_1} F=-k(x-x_0)$ The force acts in the direction opposite to that of the displacement. The constant $$k$$ is a measure of how hard or soft the spring is.

Newton’s law of motion states that the force applied on an object equals its mass multiplied by its acceleration. The variable $$h=x-x_0$$ represents the displacement of the spring from its undistorted length, and the acceleration is the second derivative of the displacement. Therefore: $\label{ode2:spring_2} F=m\frac{d^2h(t)}{dt^2}$

Combining equations \ref{ode2:spring_1} and \ref{ode2:spring_2} we obtain: $\label{ode2:spring_3} m\frac{d^2h(t)}{dt^2}=-kh(t)$

which is a second order differential equation. Notice that $$m$$ (the mass of the body) and $$k$$ (the spring constant) are not functions of $$x$$.

Assume that the displacement $$h$$ and the velocity $$h'$$ at time $$t=0$$ are: $$h(0) = A$$ and $$h'(0)=0$$. Physically, this means that the displacement at time zero is $$A$$, and the body is at rest.

$$\bullet$$ Obtain an expression for $$h(t)$$.

$$\bullet$$ What is the period of the function you found above?

In the example above we assumed that the forces due to friction were negligible. If the oscillator moves in a viscous medium, we need to include a frictional term in Newton’s equation. The force due to friction is proportional to the velocity of the mass ($$h'(t)$$), and the direction is opposite to the displacement. Therefore:

$\label{ode2:spring_4} m\frac{d^2h(t)}{dt^2}=-kh(t)-\gamma \frac{dh(t)}{dt}$

where $$\gamma$$ is a constant that depends on the viscosity of the medium.

$$\bullet$$ Obtain an expression for $$h(t)$$. You will have to consider the cases $$\gamma^2<4mk$$, $$\gamma^2=4mk$$ and $$\gamma^2>4mk$$ separately. The answers are printed below so you can check your results. Be sure you show all your work step by step.

• $$\gamma^2<4mk$$:

$h(t)=Ae^{-\gamma t/2m}\left[\cos\left(\frac{at}{2m}\right)+\frac{\gamma}{a}\sin\left(\frac{at}{2m}\right)\right], a=\sqrt{4mk-\gamma^2} \nonumber$

• $$\gamma^2=4mk$$:

$h(t)=A\left(1+\frac{\gamma}{2m}t\right)e^{-\gamma t/2m} \nonumber$

• $$\gamma^2>4mk$$:

$h(t)=\frac{A}{2}e^{-\gamma t/2m}\left[\left(e^{at/2m}+e^{-at/2m}\right)+\frac{\gamma}{a}\left(e^{at/2m}-e^{-at/2m}\right)\right], a=\sqrt{\gamma^2-4mk} \nonumber$

Problem $$\PageIndex{3}$$

Find the eigenfunctions ($$f(x)$$) and eigenvalues $$\lambda$$ of the following boundary value problems:

• $$-\frac{d^2}{dx^2}f(x)=\lambda f(x)$$, $$f(0)=0, f'(1)=0$$
• $$-\frac{d^2}{dx^2}f(x)=\lambda f(x)$$, $$f'(0)=0, f(\pi)=0$$