Chapter 15, Problem 47AP

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, the particle has its maximum speed of 20.0 m/s and is moving to the left. (a) Determine the particle’s equation of motion, specifying its position as a function of time. (b) Where in the motion is the potential energy three times the kinetic energy? (c) Find the minimum time interval required for the particle to move from x = 0 to x = 1.0 m. (d) Find the length of a simple pendulum with the same period.

To determine

The equation of the motion of the particle.

Answer to Problem 47AP

The motion equation of the motion of the particle is $x=2\cos \left(10t+\frac{\pi }{2}\right)$.

Explanation of Solution

Mass of the particle is $0.50\text{kg}$, the force constant of the spring is $50.0\text{N}/\text{m}$ and the maximum speed of the particle at time $t=0$ is $20.0\text{m}/\text{s}$.

Formula to calculate the angular frequency is,

    $\omega =\sqrt{\frac{k}{m}}$

The general equation of the particle’s motion is,

    $x=A\cos \left(\omega t+\varphi \right)$        (I)

Differentiate the above equation with respect to time.

    $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(A\cos \left(\omega t+\varphi \right)\right)\\ v=-A\omega \sin \left(\omega t+\varphi \right)\end{array}$

From the given condition, At $t=0$, $v=-v_{\max }$

Substitute these values in the above equation.

    $\begin{array}{c}-v_{\max }=-A\omega \sin \left(\omega \times 0+\varphi \right)\\ v_{\max }=A\omega \sin \varphi \end{array}$

The maximum value of $\sin \varphi$ is 1. Therefore,

    $\begin{array}{c}\sin \varphi =1\\ \varphi =\frac{\pi }{2}\end{array}$

Therefore,

    $\begin{array}{l}{v}_{\max }=A\omega \\ A=\frac{v_{\max }}{\omega }\end{array}$

Substitute $\sqrt{\frac{k}{m}}$ for $\omega$ in the above equation.

    $A=v_{\max }\sqrt{\frac{m}{k}}$

Substitute $v_{\max }\sqrt{\frac{m}{k}}$ for $A$, $\sqrt{\frac{k}{m}}$ for $\omega$ and $\frac{\pi }{2}$ for $\varphi$ in equation (I).

    $x=v_{\max }\sqrt{\frac{m}{k}}\cos \left(\sqrt{\frac{k}{m}}t+\frac{\pi }{2}\right)$

Substitute $20.0\text{m}/\text{s}$ for $v_{\max }$, $50.0\text{N}/\text{m}$ for $k$ and $0.50\text{kg}$ for $m$ in the above equation.

    $\begin{array}{c}x=20.0\text{m}/\text{s}\times \sqrt{\frac{0.50\text{kg}}{50.0\text{N}/\text{m}}}\cos \left(\sqrt{\frac{50.0\text{N}/\text{m}}{0.50\text{kg}}}t+\frac{\pi }{2}\right)\\ =2\cos \left(10t+\frac{\pi }{2}\right)\end{array}$

Here, $x$ in meter and $t$ in seconds.

Conclusion:

Therefore, the motion equation of the motion of the particle is $x=2\cos \left(10t+\frac{\pi }{2}\right)$.

To determine

The position where the potential energy is the three times the kinetic energy.

Answer to Problem 47AP

The position where the potential energy is the three times the kinetic energy is $\pm 1.73\text{m}$.

Explanation of Solution

Mass of the particle is $0.50\text{kg}$, the force constant of the spring is $50.0\text{N}/\text{m}$ and the maximum speed of the particle at time $t=0$ is $20.0\text{m}/\text{s}$.

Formula to calculate the maximum energy stored in the spring is,

    $U_{\max }=\frac{1}{2}k{A}^2$

Formula to calculate the potential energy at any position is,

    $U=\frac{1}{2}k{x}^2$

Formula to calculate the kinetic energy at any position is,

    $K=\frac{1}{2}m{v}^2$

From the given condition,

    $\frac{1}{2}m{v}^2=\frac{1}{3}\left(\frac{1}{2}k{x}^2\right)$

From the conservation of energy,

    $\begin{array}{c}U+K=U_{\max }\\ \frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2\end{array}$

Substitute $\frac{1}{3}\left(\frac{1}{2}k{x}^2\right)$ for $\frac{1}{2}m{v}^2$ in the above equation.

    $\begin{array}{c}\frac{1}{2}k{x}^2+\frac{1}{3}\left(\frac{1}{2}k{x}^2\right)=\frac{1}{2}k{A}^2\\ \frac{4}{3}{x}^2=A^2\\ x=\pm \sqrt{\frac{3}{4}}A\end{array}$

Substitute $2.0\text{m}$ for $A$ in the above equation.

    $\begin{array}{c}x=\pm \sqrt{\frac{3}{4}}\times 2.0\text{m}\\ =\pm 1.73\text{m}\end{array}$

Conclusion:

Therefore, the position where the potential energy is the three times the kinetic energy is $\pm 1.73\text{m}$.

To determine

The minimum time interval required for the particle to move from $x=0$ to $x=1.0\text{m}$.

Answer to Problem 47AP

The minimum time interval required for the particle to move from $x=0$ to $x=1.0\text{m}$ is $105\text{ms}$.

Explanation of Solution

Mass of the particle is $0.50\text{kg}$, the force constant of the spring is $50.0\text{N}/\text{m}$ and the maximum speed of the particle at time $t=0$ is $20.0\text{m}/\text{s}$.

The position of the particle is given by,

    $x=2\cos \left(10t+\frac{\pi }{2}\right)$

So, the particle will be at $x=0$ when,

    $\begin{array}{c}10t+\frac{\pi }{2}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\mathrm{...}\\ 10t=0,\pi ,2\pi ,3\pi \mathrm{...}\end{array}$        (II)

Initially, at $t=0$ the particle is at origin and moving to the left. So, next time it will be at origin when $10t=\pi$  that is moving to the right.

So, the particle will be at $x=1.0\text{m}$ when,

    $\begin{array}{c}10t+\frac{\pi }{2}=\frac{3\pi }{2}+\frac{\pi }{3}\\ 10t=\frac{11\pi }{6}-\frac{\pi }{2}\\ =\frac{4\pi }{3}\end{array}$        (III)

Therefore, the minimum time interval required for the particle to move from $x=0$ to $x=1.0\text{m}$ is,

    $\begin{array}{c}10\Delta t=\frac{4\pi }{3}-\pi \\ \Delta t=\frac{\pi }{3\times 10}\\ =0.105\text{s}\times \frac{{10}^3\text{ms}}{1\text{s}}\\ =105\text{ms}\end{array}$

Conclusion:

Therefore, the minimum time interval required for the particle to move from $x=0$ to $x=1.0\text{m}$ is $105\text{ms}$.

To determine

The length of a simple pendulum with same period.

Answer to Problem 47AP

The length of a simple pendulum with same period is $0.098\text{m}$.

Explanation of Solution

Formula to calculate the length of the pendulum is,

    $\begin{array}{c}\omega =\sqrt{\frac{g}{L}}\\ L=\frac{g}{{\omega }^2}\end{array}$

Here, $L$ is the length of the pendulum and $g$ is te acceleration due to gravity.

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $10\text{rad}/\text{s}$ for $\omega$ in the above equation.

    $\begin{array}{c}L=\frac{9.8\text{m}/{\text{s}}^{\text{2}}}{{\left(10\text{rad}/\text{s}\right)}^2}\\ =0.098\text{m}\end{array}$

Conclusion:

Therefore, the length of a simple pendulum with same period is $0.098\text{m}$.