A simple pendulum has time period T1. The point of suspension is now moved upward according to the r

Q: Why does the upward acceleration of the suspension point increase the effective gravity instead of decreasing it?

When a reference frame accelerates upwards, any object inside that frame experiences a pseudo-force in the opposite direction, which is downwards. Since gravity also acts downwards, the pseudo-force and gravity act in the same direction, thereby increasing the effective acceleration to $g_{\text{eff}} = g + a_y$.

Q: What would happen to the time period if the point of suspension was moving upwards with a constant velocity instead of a constant acceleration?

If the point of suspension moves with a constant velocity, its acceleration is zero. Since there is no acceleration, no pseudo-force acts on the bob. The effective gravity remains $g$, and the time period remains unchanged ($T_2 = T_1$).

Q: Can we solve this problem from an inertial frame of reference without using pseudo-forces?

Yes, you can solve it from the ground (inertial) frame. In this frame, the tension in the string must not only balance a component of gravity but also provide the necessary upward acceleration $a_y$ to the bob. Solving the equations of motion in this frame yields the exact same effective gravity $g_{\text{eff}} = g + a_y$.

Q: Why did we differentiate $y = kt^2$ twice?

The equation $y = kt^2$ gives the position as a function of time. The first derivative $\frac{dy}{dt}$ gives the velocity, and the second derivative $\frac{d^2y}{dt^2}$ gives the acceleration. We need the acceleration to calculate the pseudo-force acting on the pendulum bob.

Q: What if the pendulum was accelerating downwards with an acceleration equal to $g$?

If the suspension point accelerates downwards with $a_y = g$ (free fall), the effective gravity becomes $g_{\text{eff}} = g - g = 0$. In this state of weightlessness, the pendulum will not oscillate at all, and its time period becomes infinite.

Question

A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to the relation $y = kt^2$, ($k = 1	ext{ m/s}^2$), where $y$ is the vertical displacement. The time period now becomes $T_2$. The ratio of $\frac{T_1^2}{T_2^2}$ is (Take, $g = 10	ext{ m/s}^2$)

Options:

(A) $\frac{6}{5}$

(B) $\frac{5}{6}$

(C) $1$

(D) $\frac{4}{5}$

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Simple Harmonic Motion

Simple harmonic motion is one of the most beautiful and fundamental concepts in physics.

From the microscopic vibrations of atoms in a crystal lattice to the grand, sweeping oscillations of a massive pendulum, the mathematics governing these systems remains remarkably elegant.

Today, we are going to explore a classic JEE problem that tests our understanding of how a simple pendulum behaves when its frame of reference is no longer stationary.

Imagine you are standing inside an elevator holding a pendulum.

If the elevator is at rest, the pendulum swings with its standard time period.

But what happens when the elevator starts accelerating upwards?

Let's dive deep into the physics of non-inertial reference frames and uncover the solution step-by-step.

### Step 1: The Stationary Pendulum - Our Baseline

Before we introduce any motion to the point of suspension, let's establish our baseline.

For a simple pendulum of length $l$ suspended in a stationary frame, the only restoring force acting on the bob is due to gravity.

The acceleration due to gravity, $g$, acts vertically downwards.

The time period of oscillation for such a pendulum is given by the well-known formula:

$$T_1 = 2\pi \sqrt{\frac{l}{g}}$$

Here, $T_1$ represents the initial time period when the support is completely at rest.

This formula assumes small-angle oscillations, where the restoring torque is directly proportional to the angular displacement.

### Step 2: Enter Acceleration - The Moving Suspension

Now, let's look at the twist in the problem.

The point of suspension is no longer stationary; it is moving vertically upwards.

We are given that its vertical displacement $y$ varies with time $t$ according to the relation:

$$y = kt^2$$

where $k = 1	ext{ m/s}^2$.

To understand how this motion affects the pendulum, we need to find the acceleration of this suspension point.

Recall that acceleration is the second derivative of position with respect to time.

Let's differentiate $y$ once to get the velocity $v_y$:

$$v_y = \frac{dy}{dt} = 2kt$$

Now, differentiating once more with respect to time gives us the acceleration $a_y$:

$$a_y = \frac{d^2y}{dt^2} = 2k$$

Since the constant $k$ is given as $1	ext{ m/s}^2$, we can substitute this value to find the constant upward acceleration of the support:

$$a_y = 2(1) = 2	ext{ m/s}^2$$

This means our point of suspension is accelerating upwards at a constant rate of $2	ext{ m/s}^2$.

### Step 3: The Non-Inertial Frame and Pseudo-Forces

Because the point of suspension is accelerating, it constitutes a non-inertial frame of reference.

To analyze the motion of the pendulum bob from this accelerating frame, we must introduce a pseudo-force.

According to Newton's laws in non-inertial frames, any object of mass $m$ inside a frame accelerating with acceleration $\vec{a}$ experiences a pseudo-force given by:

$$\vec{F}_{	ext{pseudo}} = -m\vec{a}$$

Notice the minus sign! This indicates that the pseudo-force always acts in the direction opposite to the acceleration of the frame.

Since our suspension point is accelerating vertically upwards, the pseudo-force on the pendulum bob must act vertically downwards.

The pseudo-acceleration experienced by the bob is therefore:

$$a_{	ext{pseudo}} = 2	ext{ m/s}^2 	ext{ (downwards)}$$

### Step 4: Calculating the New Time Period

Now, let's find the effective acceleration due to gravity, $g_{	ext{eff}}$, acting on the bob.

Both the real gravitational acceleration $g$ and the pseudo-acceleration $a_{	ext{pseudo}}$ act in the same downward direction.

Therefore, we can simply add their magnitudes to find the net effective acceleration:

$$g_{	ext{eff}} = g + a_y$$

Given that $g = 10	ext{ m/s}^2$ and $a_y = 2	ext{ m/s}^2$, we get:

$$g_{	ext{eff}} = 10 + 2 = 12	ext{ m/s}^2$$

With this new effective gravity, the modified time period $T_2$ of the pendulum is:

$$T_2 = 2\pi \sqrt{\frac{l}{g_{	ext{eff}}}} = 2\pi \sqrt{\frac{l}{g + a_y}}$$

Because the effective gravity has increased, the restoring force on the bob is stronger, which means the pendulum will swing faster, resulting in a shorter time period $T_2$.

### Step 5: The Grand Finale - Finding the Ratio

We are asked to find the ratio of the squares of the time periods, $\frac{T_1^2}{T_2^2}$.

Let's write down the expressions for the squares of both time periods:

$$T_1^2 = 4\pi^2 \frac{l}{g}$$

$$T_2^2 = 4\pi^2 \frac{l}{g_{	ext{eff}}}$$

Taking the ratio of these two squared values, we see that the constants $4\pi^2$ and the length $l$ cancel out beautifully:

$$\frac{T_1^2}{T_2^2} = \frac{g_{	ext{eff}}}{g}$$

Now, substituting our calculated values of $g_{	ext{eff}} = 12	ext{ m/s}^2$ and $g = 10	ext{ m/s}^2$:

$$\frac{T_1^2}{T_2^2} = \frac{12}{10} = \frac{6}{5}$$

This elegant result of $\frac{6}{5}$ corresponds perfectly to **option (a)**.

### Beyond the Problem: What If?

To truly master physics, we should always ask "What if?" questions.

What if the point of suspension was accelerating downwards instead of upwards?

In that case, the pseudo-force would act upwards, opposing gravity, and the effective acceleration would be $g_{	ext{eff}} = g - a_y$.

This would make the pendulum swing slower, increasing its time period.

What if the support was moving with a constant velocity?

Since constant velocity means zero acceleration, there would be no pseudo-force, and the time period would remain completely unchanged!

Understanding these physical nuances is what transforms a good student into an elite JEE aspirant.

Select Answer:

Visualized Solution

Visualizing the Simple Pendulum Setup

Time Period in Stationary Frame $T_{1}$

Analyzing the Moving Suspension $y = k t^{2}$

Finding the Acceleration $a_{y}$

Introducing the Pseudo-Force Concept

Calculating Effective Gravity $g_{eff}$

New Time Period $T_{2}$

Calculating the Ratio of Time Periods

Conceptual Takeaway

The Sigma Insight: Simple Harmonic Motion (SHM)

Introduction

Step 1

Step 2

Step 3

Step 4

Step 5

Beyond the Problem

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