Chapter 5, Problem 1RE

To determine

To find: The displacement of the object whose weight is $32\text{lb}$, maximum displacement of the object from equilibrium and the time when it passes the equilibrium position first time and also how often it returns to the equilibrium position.

Answer to Problem 1RE

The displacement of the object is $\underset{\_}{x\left(t\right)=\frac{1}{3}\cos 8t}$ and the maximum displacement of the object from the equilibrium position is $\underset{\_}{\frac{1}{3}\text{ft}}$. The object passes the equilibrium position after $\underset{\_}{t=\frac{\pi }{16}\text{seconds}}$ first time and it often returns to the equilibrium position after every $\underset{\_}{\frac{\pi }{8}\text{seconds}}$.

Explanation of Solution

Procedure used:

According to Hooke’s law, a restoring force get generated in the spring if the spring is stretched a distance $s.$

That is, $F=ks$, here $k$ is spring constant.

The mass of the object is determined by the formula, $F=mg$. Here $F$ is the force, $m$ is the mass, and $g$ is gravity.

Calculation:

Since, the object weighs is $32\text{lb}$, $F=32\text{lb}$. The stretched length $s$ of the spring is $6\text{in}=\frac{6}{12}\text{ft}$. To determine the spring constant $k$, apply Hooke’s law as follows.

$\begin{array}{c}F=ks\\ 32=k\cdot \frac{6}{12}\\ k=32\cdot 2\\ =64\text{lb/ft}\end{array}$

Thus, the value of spring constant $k=64\text{lb/ft}$.

The object weighs is $32\text{lb}$, $F=32\text{lb}$ and $g=32\text{ft}/{\text{s}}^{\text{2}}$. The mass of the object is determined as follows.

$\begin{array}{l}F=mg\\ 32=m\cdot 32\\ m=1\text{slug}\end{array}$

The initial value problem that models this situation is given by the following equation.

$m\frac{d^{2}x}{d{t}^2}+kx=0$, $x\left(0\right)=\alpha ,x^{\prime }\left(0\right)=\beta$ $\left(1\right)$

Substitute the values of $m$ and $k$ in equation $\left(1\right)$.

$x^{″}+64x=0$ $\left(2\right)$

Since, at $t=0$ the object is $4\text{inches}$ below the equilibrium and it is released from the rest.

Thus, initial position $x\left(0\right)=4\text{in}$ and initial velocity is $x^{\prime }\left(0\right)=0$.

$\begin{array}{c}x\left(0\right)=4\times \frac{1}{12}\\ =\frac{1}{3}\text{ft}\end{array}$

The characteristic equation of equation $\left(2\right)$ is,

$r^2+64=0$

The solutions to the characteristic equation are as follows.

$\begin{array}{c}{r}^2=-64\\ r=64i^2\\ r=8i,-8i\end{array}$

So, the general solution of $\left(2\right)$ is $x\left(t\right)=c_1\cos 8t+c_2\sin 8t$.

Differentiate $x\left(t\right)$ with respect to $t$ as follows.

$x^{\prime }\left(t\right)=-8c_1\sin 8t+8c_2\cos 8t$

Substitute $\frac{1}{3}$ for $x\left(t\right)$ and $0$ for $t$ in $x\left(t\right)$ then

$\frac{1}{3}=c_1\cos 0+c_2\sin 0$

Thus, $c_1=\frac{1}{3}$

From $x^{\prime }\left(0\right)=0$, the following is obtained.

$\begin{array}{c}{x}^{\prime }\left(0\right)=-8c_1\sin 0+8c_2\cos 0\\ 8c_2=0\\ c_2=0\end{array}$

Thus, $c_2=0$.

Therefore the solution of the equation $\left(2\right)$ is,

$x\left(t\right)=\frac{1}{3}\cos 8t$

Thus, the maximum value $x\left(t\right)$ is $\frac{1}{3}$ at $t=0$. So the maximum displacement of the object from the equilibrium position is $\frac{1}{3}\text{ft}$.

The object will return to the equilibrium position when $x\left(t\right)=0$.

$\begin{array}{c}x\left(t\right)=\frac{1}{3}\cos 8t\\ 0=\frac{1}{3}\cos 8t\\ 8t={\cos }^{-1}\left(0\right)\\ 8t=\frac{\pi }{2}\end{array}$

Thus, the time required for the object to return to the equilibrium position is $t=\frac{\pi }{16}\text{seconds}$.

The time period of the spring-mass system is given by,

$T=2\pi \sqrt{\frac{m}{k}}$

Substitute $64$ for $k$ and $1$ for $m$ in above equation.

$\begin{array}{c}T=2\pi \sqrt{\frac{1}{64}}\\ T=\frac{2\pi }{8}\\ T=\frac{\pi }{4}\text{seconds}\end{array}$

Therefore, the object returns to the equilibrium position after every $\frac{T}{2}=\frac{\pi }{8}\text{seconds}$.