Chapter 12, Problem 74QAP

To determine

(a)

The when the object is at equilibrium

Answer to Problem 74QAP

At $t_2=0.0816\sec$ the object is at equilibrium

Explanation of Solution

Given:

Mass of the object $m=0.10kg$

Spring constant $K=15N/m$

Displacement $x_1=0.15m$

Body is released from rest $v_1=0m/s$

Formula used:

Conservation of energy
  $\frac{1}{2}K{x}_1^2+\frac{1}{2}m{v}_1^2=\frac{1}{2}K{x}_2^2+\frac{1}{2}m{v}_2^2$

Calculation:

  $\frac{1}{2}K{x}_1^2+\frac{1}{2}m{v}_1^2=\frac{1}{2}K{x}_2^2+\frac{1}{2}m{v}_2^2$

  $\frac{1}{2}K{x}_1^2=\frac{1}{2}m{v}_2^2$

  $v_2^2=\frac{K{x}_1^2}{m}$

  $\begin{array}{l}{v}_2=\sqrt{\frac{K{x}_1^2}{m}}\\ =\sqrt{\frac{(15){(0.15)}^2}{0.1}}\\ =1.837m/s\end{array}$

Speed at equilibrium point is $v_2=1.837m/s$

Time taken to reach the equilibrium position
  $\begin{array}{l}{t}_2-t_1=\Delta t=\frac{\Delta x}{\Delta v}\\ =\frac{0.15}{1.837}\\ =8.165\times {10}^{-2}\\ =0.0816\sec \end{array}$

  $\begin{array}{l}{t}_2-t_1=0.0816\sec (t_1=0)\\ t_2=0.0816\sec \end{array}$

Conclusion:

At $t_2=0.0816\sec$ the object is at equilibrium

To determine

(b)

When the object is $10.0cm$ to the left of equilibrium

Answer to Problem 74QAP

At $t_3=0.204s(t_1=0)$ the object is at equilibrium

Explanation of Solution

Given:

  $x_3=-10cm=-0.1m$

Calculation:

  $\frac{1}{2}K{x}_3^2=\frac{1}{2}m{v}_3^2$

  $v_3^2=\frac{K{x}_3^2}{m}$

  $\begin{array}{l}{v}_3=\sqrt{\frac{K{x}_3^2}{m}}\\ =\sqrt{\frac{(15){(0.1)}^2}{0.1}}\\ =1.224m/s\end{array}$

Speed at equilibrium point is $v_3=1.224m/s$

Time taken to reach the equilibrium position
  $\begin{array}{l}{t}_3-t_1=\Delta t=\frac{\Delta x}{\Delta v}\\ =\frac{-0.15-0.1}{1.224}\\ =0.204s\end{array}$

  $t_3=0.204s(t_1=0)$

Conclusion:

At $t_3=0.204s(t_1=0)$ the object is at equilibrium

To determine

(c)

When the object is $5.0cm$ to the left of equilibrium

Answer to Problem 74QAP

At $t_4=0.163s(t_1=0)$ the object is at equilibrium

Explanation of Solution

Given:

  $x_3=-5cm=-0.05m$

Calculation:

  $\frac{1}{2}K{x}_4^2=\frac{1}{2}m{v}_4^2$

  $v_4^2=\frac{K{x}_4^2}{m}$

  $\begin{array}{l}{v}_4=\sqrt{\frac{K{x}_4^2}{m}}\\ =\sqrt{\frac{(15){(0.05)}^2}{0.1}}\\ =0.612m/s\end{array}$

Speed at equilibrium point is $v_4=1.167m/s$

Time taken to reach the equilibrium position
  $\begin{array}{l}{t}_4-t_1=\Delta t=\frac{\Delta x}{\Delta v}\\ =\frac{0.1}{0.612}\\ =0.163s\end{array}$

  $t_4=0.163s(t_1=0)$

Conclusion:

At $t_4=0.163s(t_1=0)$ the object is at equilibrium