Chapter 24, Problem 6P

Review. A block having mass m and charge + Q is connected to an insulating spring having a force constant k. The block lies on a frictionless, insulating, horizontal track, and the system is immersed in a uniform electric field of magnitude E directed as shown in Figure P24.6. The block is released from rest when the spring is unstretched (at x = 0). We wish to show that the ensuing motion of the block is simple harmonic. (a) Consider the system of the block, the spring, and the electric field. Is this system isolated or nonisolated? (b) What kinds of potential energy exist within this system? (c) Call the initial configuration of the system that existing just as the block is released from rest. The final configuration is when the block momentarily comes to rest again. What is the value of x when the block comes to rest momentarily? (d) At some value of x we will call x = x₀, the block has zero net force on it. What analysis model describes the particle in this situation? (c) What is the value of x₀? (f) Define a new coordinate system x′ such that x′ = x − x₀. Show that x′ satisfies a differential equation for simple harmonic motion. (g) Find the period of the simple harmonic motion. (h) How does the period depend on the electric field magnitude?

Figure P24.6

To determine

Whether the system is isolated or non-isolated.

Answer to Problem 6P

The system is non-isolated.

Explanation of Solution

Given info: The mass of the block is $m$, charge is $+Q$, spring constant is $k$, electric field is $E$.

The system of the block with a spring is isolated system but when the system of the block with a spring is placed in a electric or magnetic field then it experiences an external force that makes the system non-isolated.

Conclusion:

Therefore, the system is non-isolated.

To determine

The kinds of potential energy exist in the system.

Answer to Problem 6P

The kinds of potential energy exist in the system are elastic potential energy and electrostatic potential energy.

Explanation of Solution

Given info: The mass of the block is $m$ , charge is $+Q$ , spring constant is $k$,electric field is $E$.

There are two kinds of potential energy exist in the system one is elastic potential energy and other is electrostatic potential energy.

The formula to calculate the elastic potential energy is,

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

Here,

$k$ is the spring constant.

$x$ is the distance of spring on stretch.

The formula to calculate the electrostatic potential energy is,

$W=Fx$ (2)

Here,

$F$ is the electrostatic force.

$x$ is the distance.

The formula to calculate electrostatic force is,

$F=QE$

Here,

$Q$ is the charge.

$E$ is the electric field.

Substitute $QE$ for $F$ in formula (2) as,

$\begin{array}{l}W=Fx\\ W=QE\end{array}$

Conclusion:

Therefore, the kinds of potential energy exist in the system are elastic potential energy and electrostatic potential energy.

To determine

The value of $x$ when the block comes to rest momentarily.

Answer to Problem 6P

The value of $x$ when the block comes to rest momentarily is $\frac{2QE}{k}$.

Explanation of Solution

Given info: The mass of the block is $m$ , charge is $+Q$ , spring constant is $k$,electric field is $E$.

The formula to calculate the elastic potential energy is,

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

Here,

$k$ is the spring constant.

$x$ is the distance of spring on stretch.

The formula to calculate the electrostatic potential energy is,

$W=Fx$ (2)

Here,

$F$ is the electrostatic force.

$x$ is the distance.

The formula to calculate electrostatic force is,

$F=QE$

Here,

$Q$ is the charge.

$E$ is the electric field.

Substitute $QE$ for $F$ in formula (2) as,

$\begin{array}{l}W=Fx\\ W=QEx\end{array}$

Equate the formula (1) and (2) as,

$\begin{array}{c}\frac{1}{2}k{x}^2=QEx\\ x=\frac{2QE}{k}\end{array}$

Conclusion:

Therefore, the value of $x$ when the block comes to rest momentarily is $\frac{2QE}{k}$.

To determine

The analysis model that describes particle in this situation.

Answer to Problem 6P

The analysis model that describes the particle in this situation is in equilibrium.

Explanation of Solution

Given info: The mass of the block is $m$ , charge is $+Q$ , spring constant is $k$,electric field is $E$.

The above statement is explained as the net force acting on the block is zero at some value of $x$ then the block said to be in equilibrium position that is the sum of the forces acting on the block balance each other.

Conclusion:

Therefore, the particle in this situation is in equilibrium.

To determine

The value of $x_0$.

Answer to Problem 6P

The value of $x_0$.is $\frac{QE}{k}$.

Explanation of Solution

Given info: The mass of the block is $m$ , charge is $+Q$ , spring constant is $k$,electric field is $E$.

The formula to calculate the spring force is,

$F_{\text{s}}=-k{x}_0$

Here,

$k$ is the spring constant.

$x$ is the displacement.

The formula to calculate electric force is,

$F_{\text{E}}=QE$

Here,

$Q$ is the electric charge.

$E$ is the electric field.

The net force is zero so it is expressed as,

$\begin{array}{l}{F}_{\text{s}}=F_{\text{E}}\\ k{x}_0=QE\end{array}$

Here,

$F_{\text{E}}$ is the electric force.

$F_{\text{s}}$ is the spring force.

Rearrange the above expression to find $x_0$,

$\begin{array}{c}k{x}_0=QE\\ x_0=\frac{QE}{k}\end{array}$

Conclusion:

Therefore, the value of $x_0$ is $\frac{QE}{k}$.

To determine

To show: The coordinate $x^{\prime }$ satisfies a differential equation for simple harmonic motion.

Answer to Problem 6P

Thus, the coordinate $x^{\prime }$ satisfies a differential equation for simple harmonic motion.

Explanation of Solution

The mass of the block is $m$ , charge is $+Q$ , spring constant is $k$,electric field is $E$,the new coordinate system $x^{\prime }$ is $x-x_0$.

Given info:

From part (e), the value of $x_0$ is $\frac{QE}{k}$.

The formula to calculate the force equation is,

${\displaystyle \sum F}=ma$ (3)

Here,

$m$ is the mass.

$a$ is the acceleration.

${\displaystyle \sum F}$ is the summation of the force.

The formula to calculate the acceleration is,

$a=\frac{d^{2}x}{d{t}^2}$

Here,

$\frac{d^{2}x}{d{t}^2}$ is the double derivative of displacement.

The summation of the force is,

$\begin{array}{c}{\displaystyle \sum F}=F_S+F_E\\ =-kx+QE\end{array}$

Here,

$F_S$ is the spring force.

$F_E$ is electrostatic force.

The new coordinate is,

$x^{\prime }=x-x_0$

Substitute $\frac{QE}{k}$ for $x_0$ in the above formula to find $x$,

$\begin{array}{c}{x}^{\prime }=x-x_0\\ x^{\prime }=x-\frac{QE}{k}\\ x=x^{\prime }+\frac{QE}{k}\end{array}$

Substitute $\frac{d^{2}x}{d{t}^2}$ for $a$ and $-kx+QE$ for ${\displaystyle \sum F}$, $x^{\prime }+\frac{QE}{k}$ for $x$ in equation (3) as,

$\frac{d^{2}x}{d{t}^2}=-kx+QE$

$\begin{array}{l}\frac{d^{2}x}{d{t}^2}=-kx+QE\\ \frac{d^2\left(x^{\prime }+\frac{QE}{k}\right)}{d{t}^2}=-kx+QE\\ \frac{d^{2}x}{d{t}^2}=-\left(\frac{k}{m}\right)x^{\prime }\end{array}$

This is the required equation of S.H.M.

Conclusion:

Therefore, the coordinate $x^{\prime }$ satisfies a differential equation for simple harmonic motion.

To determine

The period of the simple harmonic motion.

Answer to Problem 6P

The period of the simple harmonic motion is $2\pi \sqrt{\frac{m}{k}}$.

Explanation of Solution

Given info: The mass of the block is $m$ , charge is $+Q$ , spring constant is $k$,electric field is $E$,the new coordinate system $x^{\prime }$ is $x-x_0$.

From part (f), the equation for the S.H.M is,

$\frac{d^{2}x}{d{t}^2}=-\left(\frac{k}{m}\right)x^{\prime }$

The general equation of S.H.M is,

$\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$

Here,

${\omega }^2$ is the angular frequency.

Compare the equation (4) and (5) as,

$\begin{array}{c}-{\omega }^{2}x=-\left(\frac{k}{m}\right)x^{\prime }\\ {\omega }^2=\frac{k}{m}\\ \omega =\sqrt{\frac{k}{m}}\end{array}$

The formula to calculate the time period is,

$T=\frac{2\pi }{\omega }$

Here,

$\omega$ is the angular frequency.

Substitute $\sqrt{\frac{k}{m}}$ for $\omega$ in the above formula to find $T$,

$\begin{array}{c}T=\frac{2\pi }{\omega }\\ =2\pi \sqrt{\frac{m}{k}}\end{array}$

Conclusion:

Therefore, the period of the simple harmonic motion is $2\pi \sqrt{\frac{m}{k}}$.

To determine

The time period depend on the electric field magnitude or not.

Answer to Problem 6P

The time period does not depend on the electric field magnitude.

Explanation of Solution

Given info: The mass of the block is $m$ , charge is $+Q$, spring constant is $k$, electric field is $E$,the new coordinate system $x^{\prime }$ is $x-x_0$.

From part (g),the time period is,

$T=2\pi \sqrt{\frac{m}{k}}$

In the above formula, there is no term that signifies electric field magnitude. So the time period is independent of electric field magnitude.

Conclusion:

Therefore, the time period does not depend on the electric field magnitude.