Chapter 22.1, Problem 1P

A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m/s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.

To determine

The differential equation which describes the motion and the position of the block when $t=0.22\text{s}$.

Answer to Problem 1P

The differential equation which describes the motion is $\ddot{y}+56.1y=0$ and the position of the block when $t=0.22\text{s}$ is $\text{0}\text{.192}\text{m}$.

Explanation of Solution

Given:

The spring stretched, $x=175\text{mm}\text{or}\left(\text{0}\text{.175}\text{m}\right)$.

The mass of the block, $m=8\text{kg}$.

Distance moved below from equilibrium position, $y=100\text{mm}\text{or}\left(\text{0}\text{.1}\text{m}\right)$.

The downward velocity of block, $v=1.50\text{m}/\text{s}$.

The given time, $t=0.22\text{s}$.

Show the free body diagram of theblock as in Figure (1).

Conclusion:

Refer Figure (1),

Resolve forces along $y$ direction.

$\begin{array}{l}+\downarrow {\displaystyle \sum F_y}=m{a}_y\\ mg-k\left(y+y_{st}\right)=m\ddot{y}\\ mg-ky-k{y}_{st}=m\ddot{y}\end{array}$ (I)

Here, acceleration due to gravity is $g$ and spring constant is $k$.

Substitute $mg$ for $k{y}_{st}$ in Equation (I).

$\begin{array}{l}mg-ky-mg=m\ddot{y}\\ \ddot{y}+\frac{k}{m}y=0\\ \ddot{y}+{\left({\omega }_n\right)}^{2}y=0\end{array}$ (II)

Express the natural frequency.

$\begin{array}{c}{\omega }_n=\sqrt{\frac{k}{m}}\\ =\sqrt{\frac{mg/x}{m}}\end{array}$ (III)

Substitute $\text{8}\text{kg}$ for $m$, $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $\text{0}\text{.175}\text{m}$ for $x$ in Equation (III).

$\begin{array}{c}{\omega }_n=\sqrt{\frac{8\left(9.81\right)/0.175}{8}}\\ =\sqrt{56.057}\\ =7.487\end{array}$

Substitute $7.487$ for ${\omega }_n$ in Equation (II).

$\begin{array}{l}\ddot{y}+{\left(7.487\right)}^{2}y=0\\ \ddot{y}+56.1y=0\end{array}$ (IV)

Hence, the differential equation which describes the motion is $\ddot{y}+56.1y=0$.

Write the solution in Equation (IV) in the form of differential equation.

$y=A\sin {\omega }_{n}t+B\cos {\omega }_{n}t$ (V)

Here, constants are $A\text{and}B$ respectively.

Differentiate Equation (V) with respect to time to get,

$\begin{array}{c}\dot{y}=v\\ =A{\omega }_n\cos {\omega }_{n}t-B{\omega }_n\sin {\omega }_{n}t\end{array}$ (VI)

Write the boundary conditions:

At

$t=0$:

The conditions are:

$\begin{array}{l}y=0.1\text{m}\\ v=v_0\\ =1.50\text{m}/\text{s}\end{array}$

Apply the boundary conditions in Equation (V),

$\begin{array}{l}0.1=A\sin 0+B\cos 0\\ B=0.1\text{m}\end{array}$

Apply the boundary conditions in Equation (VI),

$\begin{array}{l}{v}_0=A{\omega }_n\cos 0-0\\ A=\frac{v_0}{{\omega }_n}\end{array}$ (VII)

Substitute $\text{1}\text{.50}\text{m}/\text{s}$ for $v_0$ and $7.487$ for ${\omega }_n$ in Equation (VII).

$\begin{array}{c}A=\frac{1.50}{7.487}\\ =0.2003\text{m}\end{array}$

Substitute $\text{0}\text{.2003}\text{m}$ for $A$, $7.487$ for ${\omega }_n$, $\text{0}\text{.22}\text{s}$ for $t$ and $\text{0}\text{.1}\text{m}$ for $B$ in Equation (V).

$\begin{array}{c}y=\left(0.2003\right)\sin \left[7.487\left(0.22\right)\right]+0.1\cos \left[7.487\left(0.22\right)\right]\\ =0.19971+\left(-0.00763\right)\\ =0.192\text{m}\end{array}$

Hence, the position of the block when $t=0.22\text{s}$ is $\text{0}\text{.192}\text{m}$.