Chapter 16, Problem 74PQ

The total energy of a simple harmonic oscillator with amplitude 3.00 cm is 0.500 J.

1. a. What is the kinetic energy of the system when the position of the oscillator is 0.750 cm?
2. b. What is the potential energy of the system at this position?
3. c. What is the position for which the potential energy of the system is equal to its kinetic energy?
4. d. For a simple harmonic oscillator, what, if any, are the positions for which the kinetic energy of the system exceeds the maximum potential energy of the system? Explain your answer.

FIGURE P16.73

To determine

The kinetic energy of the system.

Answer to Problem 74PQ

The kinetic energy of the system is $0.469\text{J}$ .

Explanation of Solution

Write an expression for the total energy of the system.

  $E=\frac{1}{2}k{y}_{\max }^2$                                                                                                               (I)

Here, $E$ is the total energy of the system, $k$ is the force constant and $y_{\max }$ is the maximum amplitude.

Rewrite the equation (I) to find $k$.

  $k=\frac{2E}{y_{\max }^2}$                                                                                                                (II)

Write an expression for the potential energy of the system.

  $U=\frac{1}{2}k{y}^2$                                                                                                             (III)

Here, $U$ is the potential energy and $y$ is the position.

Write an expression for the kinetic energy of the system.

  $K=E-U$                                                                                                             (IV)

Here, $K$ is the kinetic energy of the system.

Substitute equation (I) and (III) in equation (IV).

      $K=\frac{1}{2}k{y}_{\max }^2-\frac{1}{2}k{y}^2$                                                                                      (V)

Conclusion:

Substitute $0.500\text{J}$ for $E$ and $3.00\text{cm}$ for $y_{\max }$ in equation (II) to find $k$.

    $\begin{array}{c}k=\frac{2\left(0.500\text{J}\right)}{{\left(\left(3.00\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2}\\ =\frac{1.00\text{J}}{{\left(3.00\times {10}^{-2}\text{m}\right)}^2}\\ =1.11\times {10}^3\text{N/m}\end{array}$

Substitute $1.11\times {10}^3\text{N/m}$ for $k$, $3.00\text{cm}$ for $y_{\max }$ and $0.750\text{cm}$ for $y$ in equation (V) to find $K$.

  $\begin{array}{c}K=\frac{1}{2}\left(1.11\times {10}^3\text{N/m}\right){\left(\left(3.00\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2-\frac{1}{2}\left(1.11\times {10}^3\text{N/m}\right){\left(\left(0.750\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2\\ =\frac{1}{2}\left(1.11\times {10}^3\text{N/m}\right){\left(3.00\times {10}^{-2}\text{m}\right)}^2-\frac{1}{2}\left(1.11\times {10}^3\text{N/m}\right){\left(0.750\times {10}^{-2}\text{m}\right)}^2\\ =0.469\text{J}\end{array}$

Thus, the kinetic energy of the system is $0.469\text{J}$ .

To determine

The potential energy of the system.

Answer to Problem 74PQ

The potential energy of the system is $3.13\times {10}^{-2}\text{J}$.

Explanation of Solution

Write an expression for the potential energy of the system.

  $U=\frac{1}{2}k{y}^2$                                                                                                               (III)

Conclusion:

Substitute $1.11\times {10}^3\text{N/m}$ for $k$ and $0.750\text{cm}$ for $y$ in equation (III) to find $U$.

  $\begin{array}{c}U=\frac{1}{2}\left(1.11\times {10}^3\text{N/m}\right){\left(\left(0.750\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2\\ =\frac{1}{2}\left(1.11\times {10}^3\text{N/m}\right){\left(0.750\times {10}^{-2}\text{m}\right)}^2\\ =3.13\times {10}^{-2}\text{J}\end{array}$

Thus, the potential energy of the system is $3.13\times {10}^{-2}\text{J}$.

To determine

The position at which the potential energy of the system is equal to the kinetic energy.

Answer to Problem 74PQ

The position at which the potential energy of the system is equal to the kinetic energy is $2.12\text{cm}$ .

Explanation of Solution

The potential energy will be half of the total energy if the potential energy and kinetic energy are same.

Write the expression for the potential energy

  $U=\frac{1}{2}E$                                                                                                                (VI)

Substitute equation (I) and (III) in equation (VI).

    $\frac{1}{2}k{y}^2=\frac{1}{2}\left(\frac{1}{2}k{y}_{\max }^2\right)$                                                                                           (VII) 

Rewrite the equation (VII) to find $y$.

    $y=\frac{y_{\max }}{\sqrt{2}}$                                                                                                              (VIII)

Conclusion:

Substitute $3.00\text{cm}$ for $y_{\max }$ to find $y$.

    $\begin{array}{c}y=\frac{3.00\text{cm}}{\sqrt{2}}\\ =\frac{3.00\text{cm}}{1.41}\\ =2.12\text{cm}\end{array}$

Thus, the position at which the potential energy of the system is equal to the kinetic energy is $2.12\text{cm}$ .

To determine

The possibility of presence of a position for a simple harmonic oscillator at which the kinetic energy of the system exceeds the total potential energy of the system.

Answer to Problem 74PQ

No position exists for a simple harmonic oscillator at which the kinetic energy of the system exceeds the total potential energy of the system.

Explanation of Solution

The total mechanical energy is conserved for the system. The maximum potential energy is equal to the total energy of the system. The total energy of the system is the sum of kinetic energy and potential energy.

Since the total energy conserved, the total energy will be a constant. The kinetic energy can also attain a maximum that equal to the total energy. Thus, the kinetic energy will never exceed the maximum potential energy.