In the following problems, take $\mathrm{g}=32\mathrm{ft}/{\sec }^2$for the U.S. Customary System and $\mathrm{g}=9.8\mathrm{m}/{\sec }^2$for the MKS system.

A helium-filled balloon on a cord, hanging y km above a level surface, is subjected to three forces:

1. The (constant) buoyant force B exerted by the external air pressure;
2. The weight of the balloon is mg, where m is the mass of the balloon;
3. The weight of the part of the cord that has been lifted off the surface, dgy, where d is the linear density (kg/m) of the cord. (We assume that the cord is longer than the height of the balloon.)

Express Newton’s second law for the balloon and show that it is exactly analogous to the governing equation for a mass vertically hung from a spring (page 226). Find the equation of motion. What is the frequency of the balloon’s oscillations?

The general solution to (1) in the case $0<{\mathrm{b}}^2<4\mathrm{mk}$:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\left(\frac{\mathrm{b}}{2\mathrm{m}}\right)\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}\sin \left(\gamma \mathrm{t}+\mathrm{\theta }\right)$

The angular frequency:

The amplitude of the steady-state solution to equation (1) depends on the angular frequency y of the forcing function and it is given by $\mathrm{A}\left(\gamma \right)={\mathrm{F}}_0\mathrm{M}\left(\gamma \right)$, where

(13) $\mathrm{M}\left(\gamma \right):=\frac{1}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}$ … (1)

The undamped system:

The system is governed by $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \gamma \mathrm{t}$. And the homogenous solution of it is ${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right),\omega :=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$. And the corresponding homogeneous equation is (21) ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$. And the correct form is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcos}\omega \mathrm{t}+{\mathrm{A}}_2\mathrm{tsin}\omega \mathrm{t}$.

So, the general solution of the system is $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$.

Newton’s second law:

The forces, $\mathrm{F}=\mathrm{ma}$includes the buoyant force B pointing up, the weight mg pointing down, and the cord weight dgy pointing down.

Given that, if y (t) is the height of the balloon at time t, Newton’s second law becomes $\mathrm{B}-\mathrm{mg}-\mathrm{dgy}=\mathrm{my}$.

If we rewrite the above equation as $\mathrm{my}+\mathrm{dgy}=\mathrm{mg}-\mathrm{B}$. Then, this equation becomes completely analogous to the equation for a mass vertically hung from a spring.

Then, $\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky}=\mathrm{mg}+{\mathrm{F}}_{\mathrm{ext}}\left(\mathrm{t}\right)$.

Let us take, b = 0, k = dg and ${\mathrm{F}}_{\mathrm{ext}}=-\mathrm{B}$. We get $\mathrm{my}+\mathrm{dgy}=\mathrm{mg}-\mathrm{B}$, …… (2)

Now find the roots of the equation.

The auxiliary equation is ${\mathrm{mr}}^2+\mathrm{dg}=0$. And its roots are $\mathrm{r}=\pm \mathrm{i}\sqrt{\frac{\mathrm{dg}}{\mathrm{m}}}$.

Then, the homogeneous equation is ${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{\frac{\mathrm{dg}}{\mathrm{m}}}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\sqrt{\frac{\mathrm{dg}}{\mathrm{m}}}\mathrm{t}\right)$.

And the particular solution in the form of constant, ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{A}$.

Let’s plug it into equation (2).

$$\begin{gathered} \mathrm{m}\left(\mathrm{A}\right)+\mathrm{dg}\left(\mathrm{A}\right)=\mathrm{mg}-\mathrm{B} \\ \mathrm{A}=\frac{\mathrm{mg}-\mathrm{B}}{\mathrm{dg}} \end{gathered}$$

Then, ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{\mathrm{mg}-\mathrm{B}}{\mathrm{dg}}$.

So, the solution of the balloon equation is:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{\frac{\mathrm{dg}}{\mathrm{m}}}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\sqrt{\frac{\mathrm{dg}}{\mathrm{m}}}\mathrm{t}\right)+\frac{\mathrm{mg}-\mathrm{B}}{\mathrm{dg}}$

Now find the period of motion.

$$\begin{gathered} \mathrm{T}=\frac{2\mathrm{\pi }}{\sqrt{\frac{\mathrm{dg}}{\mathrm{m}}}} \\ =\frac{2\mathrm{\pi }\sqrt{\mathrm{m}}}{\sqrt{\mathrm{dg}}} \end{gathered}$$

And the frequency is:

$$\begin{gathered} \mathrm{f}=\frac{1}{\mathrm{T}} \\ =\frac{\sqrt{\mathrm{dg}}}{2\mathrm{\pi }\sqrt{\mathrm{m}}} \end{gathered}$$