Chapter 12A, Problem 1PLA

To determine

The spring potential energy.

Answer to Problem 1PLA

Option (c) Maximum of $\frac{1}{2}k{A}^2$ at $y=y_0+A$ and $y=y_0-A$.

Explanation of Solution

Write the formula to find spring potential energy

    $US=\frac{1}{2}k{\left(y-y_0\right)}^2$        (I)

Here, $US$ is the spring potential energy, $k$ is the spring constant, $y$ is the stretched length of the spring and $y_0$ is the equilibrium positon.

Conclusion:

1. (a) Substitute $y_0+A$ for $y$ in equation (I)

  $\begin{array}{c}US=\frac{1}{2}k{\left(y_0+A-y_0\right)}^2\\ =\frac{1}{2}k{A}^2\end{array}$

Thus, option (a) is incorrect.

1. (b) Substitute $y_0-A$ for $y$ in equation (I)

  $\begin{array}{c}US=\frac{1}{2}k{\left(y_0-A-y_0\right)}^2\\ =\frac{1}{2}k{A}^2\end{array}$

Thus, option (b) is incorrect.

1. (c) From above calculations, it is true that the maximum of $\frac{1}{2}k{A}^2$ at $y=y_0+A$ and $y=y_0-A$.

Thus, option (c) is correct.

1. (d) Substitute $y_0$ for $y$ in equation (I)

  $\begin{array}{c}US=\frac{1}{2}k{\left(y_0-y_0\right)}^2\\ =0\end{array}$

Thus, option (d) is incorrect.