Chapter 22, Problem 81P

a.

To determine

The mass of the particle.

The mass of the particle is $m=0.997\text{kg}$

Given:

The charge of the ring $Q=5\mu \text{C}$

The charge of the particle $\text{q=-5}\mu \text{C}$

The radius of the ring $\text{a=8}\text{.0cm}$

The frequency of oscillation $f=3.34\text{Hz}$

Also, that the displacement is much small than the radius of the ring $\text{x<<a}$

Formula Used:

Electric field by a ring as a function of x.

  $E=\frac{kqx}{{(x^2+a^2)}^{\frac{3}{2}}}$

E is the electric field.

k is a constant.

q is the charge of the ring.

a is the radius of the ring.

x is the distance from center of the ring.

Calculations:

  $E=\frac{kQx}{{(x^2+a^2)}^{\frac{3}{2}}}$

  $E=\frac{kQx}{{[a^2(1+\frac{x^2}{a^2})]}^{\frac{3}{2}}}$

As $\text{x<<a}$

  $\frac{x^2}{a^2}\approx 0$

  $E\approx \frac{kQ}{a^3}x$

Now force is

  $F=qE=\frac{kqQ}{a^3}x$

As the negatively charged particle experiences a restoring force, the motion will be a simple harmonic motion.

  $F=-\frac{kqQ}{a^3}x$

  $F=m\frac{d^{2}x}{d{t}^2}$

Equating

  $-\frac{kqQ}{a^3}x=m\frac{d^{2}x}{d{t}^2}$

  $\frac{d^{2}x}{d{t}^2}+\frac{kqQ}{m{a}^3}x=0$

Relating with the acceleration of a particle executing a simple harmonic motion.

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

This is the differential equation of a simple harmonic motion.

Hence

  $\omega =\sqrt{\frac{kqQ}{m{a}^3}}$

Solving for m

  $m=\frac{kqQ}{{\omega }^2{a}^3}$

  $\omega =2\pi f$

  $m=\frac{kqQ}{4{\pi }^2{f}^2{a}^3}$

Substituting the values in the equation.

  $\begin{array}{l}m=\frac{(8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2)(-5\mu \text{C)(}5\mu \text{C})}{4{\pi }^2{(3.34\text{Hz})}^2{(.08\text{m})}^3}\\ m=0.997\text{kg}\end{array}$

Conclusion:

The mass of the particle is $m=0.997\text{kg}$

b.

The frequency of the motion if the radius of the ring is doubled.

The frequency is $f=1.2\text{Hz}$ .

Given:

The frequency is doubled.

Formula Used:

Angular frequency is

  $\omega =\sqrt{\frac{kqQ}{m{a}^3}}$

k is a constant.

q is the charge of the ring.

a is the radius of the ring.

x is the distance from center of the ring.

Calculations:

  $\omega =\sqrt{\frac{kqQ}{m{a}^3}}$

Now comparing the angular frequency when the radius is doubled.

  $\frac{\omega \text{'}}{\omega }=\frac{\sqrt{\frac{kqQ}{m{(2a)}^3}}}{\sqrt{\frac{kqQ}{m{(a)}^3}}}=\frac{1}{\sqrt{8}}$

  $\frac{\omega \text{'}}{\omega }=\frac{2\pi f\text{'}}{2\pi f}=\frac{f\text{'}}{f}$

  $\frac{f\text{'}}{f}=\frac{1}{\sqrt{8}}$

  $f\text{'}=\frac{f}{\sqrt{8}}$

  $f\text{'}=\frac{3.4\text{Hz}}{\sqrt{8}}$

  $f\text{'}=1.2\text{Hz}$

Conclusion:

The frequency is $f=1.2\text{Hz}$ .

b.

To determine

The frequency of the motion if the radius of the ring is doubled.

The frequency is $f=1.2\text{Hz}$ .

Given:

The frequency is doubled.

Formula Used:

Angular frequency is

  $\omega =\sqrt{\frac{kqQ}{m{a}^3}}$

k is a constant.

q is the charge of the ring.

a is the radius of the ring.

x is the distance from center of the ring.

Calculations:

  $\omega =\sqrt{\frac{kqQ}{m{a}^3}}$

Now comparing the angular frequency when the radius is doubled.

  $\frac{\omega \text{'}}{\omega }=\frac{\sqrt{\frac{kqQ}{m{(2a)}^3}}}{\sqrt{\frac{kqQ}{m{(a)}^3}}}=\frac{1}{\sqrt{8}}$

  $\frac{\omega \text{'}}{\omega }=\frac{2\pi f\text{'}}{2\pi f}=\frac{f\text{'}}{f}$

  $\frac{f\text{'}}{f}=\frac{1}{\sqrt{8}}$

  $f\text{'}=\frac{f}{\sqrt{8}}$

  $f\text{'}=\frac{3.4\text{Hz}}{\sqrt{8}}$

  $f\text{'}=1.2\text{Hz}$

Conclusion:

The frequency is $f=1.2\text{Hz}$ .