Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 22, Problem 81P

a.

To determine

The mass of the particle.

The mass of the particle is m=0.997kg

Given:

The charge of the ring Q=5μC

The charge of the particle q=-5μC

The radius of the ring a=8.0cm

The frequency of oscillation f=3.34Hz

Also, that the displacement is much small than the radius of the ring x<<a

Formula Used:

Electric field by a ring as a function of x.

  E=kqx(x2+a2)32

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=kQx(x2+a2)32

  E=kQx[a2(1+x2a2)]32

As x<<a

  x2a20

  EkQa3x

Now force is

  F=qE=kqQa3x

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

  F=kqQa3x

  F=md2xdt2

Equating

  kqQa3x=md2xdt2

  d2xdt2+kqQma3x=0

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

  d2xdt2=ω2x(t)

This is the differential equation of a simple harmonic motion.

Hence

  ω=kqQma3

Solving for m

  m=kqQω2a3

  ω=2πf

  m=kqQ4π2f2a3

Substituting the values in the equation.

  m=(8.988×109Nm2/C2)(5μC)(5μC)4π2(3.34Hz)2(.08m)3m=0.997kg

Conclusion:

The mass of the particle is m=0.997kg

b.

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

The frequency is f=1.2Hz .

Given:

The frequency is doubled.

Formula Used:

Angular frequency is

  ω=kqQma3

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:

  ω=kqQma3

Now comparing the angular frequency when the radius is doubled.

  ω'ω=kqQm(2a)3kqQm(a)3=18

  ω'ω=2πf'2πf=f'f

  f'f=18

  f'=f8

  f'=3.4Hz8

  f'=1.2Hz

Conclusion:

The frequency is f=1.2Hz .

b.

To determine

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

The frequency is f=1.2Hz .

Given:

The frequency is doubled.

Formula Used:

Angular frequency is

  ω=kqQma3

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:

  ω=kqQma3

Now comparing the angular frequency when the radius is doubled.

  ω'ω=kqQm(2a)3kqQm(a)3=18

  ω'ω=2πf'2πf=f'f

  f'f=18

  f'=f8

  f'=3.4Hz8

  f'=1.2Hz

Conclusion:

The frequency is f=1.2Hz .

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