A thin fixed ring of radius 1 m has a positive charge 1 x 10^-5 C uniformly distributed over it. A p

Q: Why can we ignore $x^2$ in the denominator but not the $x$ in the numerator?

This is a classic approximation concept! In the denominator, we are adding a very small number ($x^2 = 0.0001\text{ m}^2$) to a much larger number ($R^2 = 1\text{ m}^2$). Adding $0.0001$ to $1$ makes virtually no difference, so $R^2 + x^2 \approx R^2$. However, in the numerator, $x$ is a multiplying factor. If we set $x = 0$ in the numerator, the entire force becomes zero, which destroys the physical information about the displacement. Therefore, we keep the leading-order term in $x$ to find the restoring force.

Q: What happens if the charge of the particle is positive instead of negative?

If the particle had a positive charge $+q$, the force would be repulsive, pointing away from the center: $F_e = +k_{\text{eff}}x$. Instead of pulling the particle back, this force would push it further away. The equilibrium at the center would become unstable, and the particle would accelerate away to infinity instead of oscillating.

Q: Is the motion perfectly simple harmonic if $x$ is not small?

No, it is not. The true force is $F_e \propto \frac{x}{(R^2 + x^2)^{3/2}}$, which is non-linear. Perfect simple harmonic motion requires the restoring force to be strictly linear with displacement ($F \propto -x$). For larger displacements, the motion is still periodic, but it is no longer simple harmonic.

Q: Why did we convert the mass from grams to kilograms?

In physics calculations, we must always use standard SI units to ensure consistency. The unit of force (Newton) and the electrostatic constant ($1/4\pi\varepsilon_0$) are defined in terms of kilograms, meters, and seconds. Therefore, converting $0.9\text{ g}$ to $9 \times 10^{-4}\text{ kg}$ is absolutely necessary to get the correct answer in seconds.

Q: What is the velocity of the particle when it passes through the center of the ring?

When the particle passes through the center ($x = 0$), its potential energy is minimum (zero) and its kinetic energy is maximum. Its velocity at this point is the maximum velocity of SHM, given by $v_{\text{max}} = \omega A$, where $\omega = \sqrt{k_{\text{eff}}/m}$ is the angular frequency and $A$ is the amplitude of oscillation ($1\text{ cm}$).

Question

A thin fixed ring of radius $1	ext{ m}$ has a positive charge $1 	imes 10^{-5}	ext{ C}$ uniformly distributed over it. A particle of mass $0.9	ext{ g}$ and having a negative charge of $1 	imes 10^{-6}	ext{ C}$ is placed on the axis at a distance of $1	ext{ cm}$ from the centre of the ring. Show that the motion of the negatively charged particle is approximately simple harmonic. Calculate the time period of oscillations.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a beautifully symmetric physical system: a thin, fixed ring of radius $R = 1	ext{ m}$ carrying a total positive charge $Q = 10^{-5}	ext{ C}$ distributed uniformly along its circumference.

Now, let's place a tiny particle of mass $m = 0.9	ext{ g}$ and negative charge $-q = -10^{-6}	ext{ C}$ on the axis of this ring at a very small distance $x = 1	ext{ cm}$ from the center $O$.

Because the ring is positively charged and the particle is negatively charged, there is an attractive electrostatic force pulling the particle back toward the center of the ring.

Our goal is to show that for small displacements, this electrostatic force behaves exactly like a mechanical spring, leading to simple harmonic motion (SHM), and then calculate the time period of these oscillations.

### The Master Equation

To find the force acting on the particle, we first need to determine the electric field produced by the ring at any point on its axis.

Using Coulomb's Law and integrating over the ring's circumference, the electric field $E$ at a distance $x$ along the axis is given by the standard formula:

$$E = \frac{1}{4\pi\varepsilon_0} \frac{Qx}{(R^2 + x^2)^{3/2}}$$

This electric field points radially outward along the axis of the ring.

Since our particle has a negative charge $-q$, the electrostatic force $\vec{F}_e$ acting on it is:

$$F_e = -qE = -\frac{1}{4\pi\varepsilon_0} \frac{Qqx}{(R^2 + x^2)^{3/2}}$$

The negative sign mathematically represents that this is a **restoring force**, always acting in the direction opposite to the displacement $x$, pulling the particle back toward the center $O$.

### The Power of Approximation

Here comes the crucial pedagogical breakthrough.

The problem states that the particle is placed at a distance of just $1	ext{ cm}$ ($0.01	ext{ m}$), while the radius of the ring is a massive $1	ext{ m}$.

Since the displacement $x$ is much, much smaller than the radius $R$ ($x \ll R$), we can safely neglect $x^2$ in comparison to $R^2$ in the denominator:

$$R^2 + x^2 \approx R^2$$

Raising this to the power of $3/2$ simplifies the denominator to:

$$(R^2 + x^2)^{3/2} \approx (R^2)^{3/2} = R^3$$

Substituting this approximation back into our force equation yields:

$$F_e \approx -\left( \frac{Qq}{4\pi\varepsilon_0 R^3} ight) x$$

Notice the elegance of this result!

All the terms inside the parentheses are constants.

This means the restoring force is directly proportional to the displacement $x$ and acts in the opposite direction:

$$F_e \propto -x$$

This is the exact mathematical definition of **Simple Harmonic Motion**!

Thus, we have successfully shown that the particle executes SHM.

### Finding the Time Period

In simple harmonic motion, the restoring force is written as $F_e = -k_{	ext{eff}} x$, where $k_{	ext{eff}}$ is the effective force constant of the system.

By comparing this with our simplified force equation, we identify the effective force constant as:

$$k_{	ext{eff}} = \frac{Qq}{4\pi\varepsilon_0 R^3}$$

The time period $T$ of a simple harmonic oscillator is given by the classic formula:

$$T = 2\pi \sqrt{\frac{m}{k_{	ext{eff}}}}$$

Substituting our expression for $k_{	ext{eff}}$ into this formula gives:

$$T = 2\pi \sqrt{\frac{4\pi\varepsilon_0 m R^3}{Qq}}$$

### Final Calculation

Now, let's carefully substitute the given numerical values into our time period equation:

*   $Q = 10^{-5}	ext{ C}$
*   $q = 10^{-6}	ext{ C}$
*   $R = 1	ext{ m}$
*   $m = 0.9	ext{ g} = 9 	imes 10^{-4}	ext{ kg}$
*   $4\pi\varepsilon_0 = \frac{1}{9 	imes 10^9}	ext{ C}^2/(	ext{N m}^2)$

Plugging these in:

$$T = 2\pi \sqrt{\frac{(9 	imes 10^{-4}) 	imes 1^3}{(9 	imes 10^9) 	imes 10^{-5} 	imes 10^{-6}}}$$

Let's simplify the expression inside the square root step-by-step:

1.  Simplify the denominator: 
    $$(9 	imes 10^9) 	imes 10^{-11} = 9 	imes 10^{-2}$$
2.  Divide the numerator by the denominator:
    $$\frac{9 	imes 10^{-4}}{9 	imes 10^{-2}} = 10^{-2}$$
3.  Take the square root:
    $$\sqrt{10^{-2}} = 10^{-1} = 0.1$$

Thus, the time period is:

$$T = 2\pi 	imes 0.1 = 0.2\pi \approx 0.628	ext{ s}$$

This beautiful result shows that the particle will oscillate back and forth through the center of the ring with a time period of approximately **$0.628	ext{ s}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Electric Field on the Axis of a Ring

Electrostatic Force on the Particle

Applying the Small Displacement Approximation

Proving Simple Harmonic Motion

Identifying the Effective Force Constant

Formula for the Time Period

Substituting the Numerical Values

Final Calculation

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Master Equation

The Power of Approximation

Finding the Time Period

Final Calculation

Similar Questions

Starting from the origin, a body oscillates simple harmonically with a period of $2 s$ . After what time will its kinetic energy be $75%$ of the total energy?

Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of spring constant $k$ as shown in the figure. When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $m_{2}$ .

A particle of mass $m$ is executing oscillation about the origin on the $X$ -axis. Its potential energy is $U (x) = k ∣ x ∣^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is $a$ , then its time period $T$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Electric Field on the Axis of a Ring

Electrostatic Force on the Particle

Applying the Small Displacement Approximation

Proving Simple Harmonic Motion

Identifying the Effective Force Constant

Formula for the Time Period

Substituting the Numerical Values

Final Calculation

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Master Equation

The Power of Approximation

Finding the Time Period

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy?

Starting from the origin, a body oscillates simple harmonically with a period of $2 s$ . After what time will its kinetic energy be $75%$ of the total energy?