A simple pendulum of length L is placed between the plates of a parallel plate capacitor having elec

Q: Why do we use vector addition for the forces?

Force is a vector quantity. Since gravity acts downwards and the electric force acts horizontally, they are perpendicular. We must use vector addition (Pythagorean theorem in this case) to find the true net force.

Q: What exactly is 'effective gravity'?

Effective gravity ($g_{\text{eff}}$) is the net acceleration experienced by the bob due to all external non-contact forces (like gravity, electric forces, or pseudo forces). It replaces $g$ in the standard pendulum formula.

Q: Does the tension in the string affect the time period?

Tension is a constraint force that keeps the bob moving in a circular arc. It provides the necessary centripetal force and balances the radial components of other forces, but the restoring force that determines the time period depends only on the external field forces (gravity and electric field).

Q: What if the charge on the bob was negative?

If the charge was negative, the electric force would point in the opposite direction (to the left instead of right). However, since the forces are still perpendicular, the magnitude of the net force and the effective acceleration would remain exactly the same.

Q: Will this pendulum hang straight down when at rest?

No. Because of the horizontal electric force, the new equilibrium position will be shifted to an angle $\theta$ from the vertical, where $\tan\theta = \frac{qE}{mg}$. The pendulum will oscillate around this new diagonal equilibrium position.

Question

A simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having electric field $E$, as shown in figure. Its bob has mass $m$ and charge $q$. The time period of the pendulum is given by

Options:

(A) $2\pi \sqrt{\frac{L}{\sqrt{g^2 + \left(\frac{qE}{m}\right)^2}}}$

(B) $2\pi \sqrt{\frac{L}{\sqrt{g^2 - \frac{q^2E^2}{m^2}}}}$

(C) $2\pi \sqrt{\frac{L}{\left(g + \frac{qE}{m}\right)}}$

(D) $2\pi \sqrt{\frac{L}{\left(g - \frac{qE}{m}\right)}}$

Sigma Percentile · Accepted Answer

### The Concept of Effective Gravity

When a simple pendulum oscillates in a standard environment, the only non-contact force acting on the bob is gravity. The restoring force that brings the pendulum back to its mean position is a component of this gravitational force. The time period is elegantly given by the formula:

$$T = 2\pi \sqrt{\frac{L}{g}}$$

But what happens when we introduce a new player into the game? In this problem, the pendulum bob carries a charge $q$ and is placed in a uniform horizontal electric field $E$. This means the bob now experiences an electrostatic force in addition to gravity.

### Analyzing the Forces

Let's break down the forces acting on the bob. We have the standard weight acting vertically downwards:

$$F_g = mg$$

Simultaneously, the electric field exerts a horizontal force on the charged bob:

$$F_e = qE$$

Because these two forces are perpendicular to each other, they don't simply add up algebraically. We must find their vector sum to determine the net force acting on the bob. Using the Pythagorean theorem, the magnitude of this net force is:

$$F_{	ext{net}} = \sqrt{(mg)^2 + (qE)^2}$$

### Calculating the Effective Acceleration

This net force acts like a new, stronger 'gravity' pulling the bob diagonally. To use our standard time period formula, we need to find the effective acceleration, often denoted as $g_{	ext{eff}}$ or $a_{	ext{net}}$. By Newton's second law, acceleration is force divided by mass:

$$a_{	ext{net}} = \frac{F_{	ext{net}}}{m} = \frac{\sqrt{(mg)^2 + (qE)^2}}{m}$$

Bringing the mass $m$ inside the square root, we get:

$$a_{	ext{net}} = \sqrt{g^2 + \left(\frac{qE}{m}ight)^2}$$

### The Final Time Period

Now, we simply substitute this effective acceleration back into the standard time period formula. The pendulum behaves exactly like a normal pendulum, but in a world where gravity is slightly stronger and acts at an angle.

$$T = 2\pi \sqrt{\frac{L}{a_{	ext{net}}}}$$

Substituting our expression for $a_{	ext{net}}$, we arrive at the final answer:

**$$T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + \left(\frac{qE}{m}ight)^2}}}$$**

This powerful concept of 'effective gravity' can be applied to many situations, such as a pendulum in an accelerating elevator or a car taking a turn. Always identify the non-contact forces, find their resultant, and calculate the effective acceleration!

Animated Solution for Physics - Oscillations: A simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having electric field $E$ , as shown in figure. Its bob has mass $m$ and charge $q$ . The time period of the pendulum is given by

Select Answer:

Visualized Solution

Visualizing the Setup

Free Body Diagram

Net Force

Effective Acceleration

Time Period Formula

The Way Forward

The Sigma Insight: Simple Harmonic Motion (SHM)

The Concept of Effective Gravity

Analyzing the Forces

Calculating the Effective Acceleration

The Final Time Period

Similar Questions

A pendulum made of a uniform wire of cross-sectional area $A$ has time period $T$ . When an additional mass $M$ is added to its bob, the time period changes $T_{M}$ . If the Young's modulus of the material of the wire is $Y$ , then $\frac{1}{Y}$ is equal to ( $g =$ gravitational acceleration)

A simple pendulum oscillating in air has period $T$ . The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}$ th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is

The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is

Time period of a simple pendulum is $T$ . The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{α}{β} T$ . The value of $α$ is ......... .

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Time period of a simple pendulum is $T$ inside a lift, when the lift is stationary. If the lift moves upwards with an acceleration $g /2$ , then the time period of pendulum will be