Q10 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

Show that the period of the simple harmonic motion of a mass hanging from a spring is $2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$, where l denotes the amount (beyond its natural length) that the spring is stretched when the mass is at equilibrium.

Step-by-Step Solution

Answer

Thus, it is proved that the simple harmonic motion of a mass hanging from a spring is $2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$.

Step 1: General form

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

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

 

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 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$. And the correct form is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcos}\omega \mathrm{t}+{\mathrm{A}}_2\mathrm{tsin}\omega \mathrm{t}$.

Step 2: Evaluate the equation

Referring to Hooke’s law:

 

It gives us the spring constant k.

$\mathrm{l}=\frac{\mathrm{mg}}{\mathrm{k}}\Rightarrow \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{l}}$

Let l be the amount the spring is stretched.

The equation of the simple harmonic motion is $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}=0$.

This can be rewritten as:

$$\begin{gathered} \mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\frac{\mathrm{mg}}{\mathrm{l}}\mathrm{y}=0 \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\frac{\mathrm{g}}{\mathrm{l}}\mathrm{y}=0 \end{gathered}$$

Then, the auxiliary equation is ${\mathrm{r}}^2+\frac{\mathrm{g}}{\mathrm{l}}=0$ and its roots are $\mathrm{r}=\pm \sqrt{\frac{\mathrm{g}}{\mathrm{l}}}\mathrm{i}$.

And the equation of motion is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{\frac{\mathrm{g}}{\mathrm{l}}}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\sqrt{\frac{\mathrm{g}}{\mathrm{l}}}\mathrm{t}\right)$

 

From the above equation, the period must be:

 $$\begin{gathered} \mathrm{p}=\frac{2\mathrm{\pi }}{\sqrt{\frac{\mathrm{g}}{\mathrm{l}}}} \\ =2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}} \end{gathered}$$

Thus, the period of the simple harmonic motion is:

$\mathrm{p}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$