Chapter 12, Problem 51QAP

To determine

(a)

To plot:

The graph of potential energy versus time for three full periods of motion.

Answer to Problem 51QAP

The plot of potential energy is as below:

  

Explanation of Solution

Given data:

The potential energy of a simple harmonic oscillator, $U(t)=\frac{1}{2}k{x}^2$. Also, the equation of motion for a given simple harmonic oscillator is $x(t)=A\cos (\omega t)$

Calculation:

The equation of motion for a given simple harmonic oscillator is $x(t)=A\cos (\omega t)$.

The potential energy of a simple harmonic oscillator is given by $U(t)=\frac{1}{2}k{x}^2$

The plots of U ( t ) should look like those for ${\cos }^2(\omega t)$.

  $U(t)=\frac{1}{2}k{x}^2=\frac{1}{2}k{\left[A\cos (\omega t)\right]}^2=\frac{1}{2}k{A}^2{\cos }^2(\omega t)$

For simplicity, the vertical axis of the following plot is $\frac{U}{\left(\frac{1}{2}k{A}^2\right)}$, and the horizontal axis is $\omega t$ .The plot is like a cosine function as below:

  

Figure 1

Conclusion:

Thus, the plot of potential energy versus time for three full periods of motion is shown in Figure 1.

To determine

(b)

The expression for the velocity, $v(t)$.

Answer to Problem 51QAP

The expression for the velocity is given by, $v_x(t)=-\omega A\sin (\omega t)$.

Explanation of Solution

Given data:

The potential energy of a simple harmonic oscillator, $U(t)=\frac{1}{2}k{x}^2$

Calculation:

The equation of motion for a given simple harmonic oscillator is $x(t)=A\cos (\omega t)$.

The velocity of this oscillator is given by Equation,

  $\begin{array}{l}{v}_x=\frac{dx(t)}{d(t)}\left[A\cos (\omega t)\right]\\ v_x=-\omega A\sin (\omega t)\end{array}$

Conclusion:

Thus, expression for the velocity, $v_x(t)=-\omega A\sin (\omega t)$

To determine

(c)

To plot:

The graph of Kinetic energy on the same graph where potential energy versus time is drawn.

Explanation of Solution

Given data:

The potential energy of a simple harmonic oscillator, $U(t)=\frac{1}{2}k{x}^2$

Calculation:

The equation of motion for a given simple harmonic oscillator is $x(t)=A\cos (\omega t)$.

The velocity of this oscillator is given by equation, $v_x=-\omega A\sin (\omega t)$.

The potential energy of a simple harmonic oscillator is given by $U(t)=\frac{1}{2}k{x}^2$

And the kinetic energy is given by $K(t)=\frac{1}{2}m{v}_x{}^2$

The plots of U ( t ) and K ( t ) should look like those for ${\cos }^2(\omega t)$ and ${\sin }^2(\omega t)$ respectively.

  $K(t)=\frac{1}{2}m{v}_x{}^2=\frac{1}{2}m{\left[-\omega A\cos (\omega t)\right]}^2=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2(\omega t)=\frac{1}{2}k{A}^2{\sin }^2(\omega t)$

For simplicity, the vertical axis of the following plot is $\frac{U}{\left(\frac{1}{2}k{A}^2\right)},\frac{K}{\left(\frac{1}{2}k{A}^2\right)}$, and the horizontal axis is $\omega t$. The plot is similar to a Sinusoidal function as below shown with red color: -

  

Figure 2

Conclusion:

Thus, the plot of kinetic energy versus time and potential energy versus time is shown in Figure 2.