A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a height R a

Q: Why can't we use the approximation formula $g' = g(1 - 2h/R)$ here?

The approximation formula $g' \approx g(1 - \frac{2h}{R})$ is derived using a binomial expansion that is only valid when the height $h$ is extremely small compared to the Earth's radius ($h \ll R$). In this problem, $h = R$, which is definitely not small! Therefore, we must use the exact formula $g' = \frac{g}{(1 + h/R)^2}$ to avoid massive errors.

Q: Does the mass of the pendulum bob affect the time period at all?

No, the mass of the bob does not affect the time period. As seen in the formula $T = 2\pi\sqrt{\frac{L}{g'}}$, the time period depends only on the length of the string $L$ and the local gravitational acceleration $g'$. This is a beautiful consequence of the equivalence principle—all masses fall with the same acceleration under gravity.

Q: If gravity decreases, shouldn't the pendulum swing faster because it is lighter?

Actually, it's the opposite! Gravity provides the restoring force that pulls the pendulum bob back to its equilibrium position. When gravity is weaker, this restoring force is smaller, meaning the bob accelerates more slowly. As a result, it takes longer to complete a swing, which increases the time period.

Q: What does a time period ratio of 2 actually mean for a clock?

A ratio of $\frac{T_2}{T_1} = 2$ means that the pendulum takes twice as long to complete one full swing at height $R$ compared to the surface. If this pendulum were used to run a grandfather clock, the clock would run at half-speed, losing 12 hours every day!

Q: What would happen to the time period if we took the pendulum to infinite height?

As the height $h$ approaches infinity, the local gravity $g'$ approaches zero. Looking at our formula $T = 2\pi\sqrt{\frac{L}{g'}}$, as $g'$ goes to zero, the time period $T$ approaches infinity. Physically, this means the pendulum would never return to the center once displaced—it would stop swinging entirely!

Question

A simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface, where $R$ is the radius of the earth. The value of $\frac{T_2}{T_1}$ is

Options:

(A) $1$

(B) $\sqrt{2}$

(C) $4$

(D) $2$

Sigma Percentile · Accepted Answer

### Introduction to the Pendulum and Gravity

Imagine holding a simple pendulum in your hand.

As you release it, it swings back and forth with a comforting, rhythmic regularity.

This rhythm is governed by a beautiful dance between the length of the string and the invisible pull of gravity beneath your feet.

But what happens if we take this pendulum on an adventure?

What if we transport it high above the Earth's surface, where the planet's gravitational grip begins to weaken?

In this problem, we explore exactly how the time period of a simple pendulum changes when it is lifted to a height equal to the Earth's radius, $R$.

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### The Physics of the Pendulum

To understand this behavior, we must first look at the master equation governing a simple pendulum's motion.

For small oscillations, the time period $T$ is given by the classic formula:

$$T = 2\pi \sqrt{\frac{L}{g'}}$$

Here, $L$ represents the length of the pendulum, and $g'$ is the local acceleration due to gravity.

Since we are using the same pendulum in both locations, the length $L$ remains strictly constant.

This reveals a fundamental proportionality:

$$T \propto \frac{1}{\sqrt{g'}}$$

This inverse-square-root relationship tells us that if gravity weakens, the time period must increase.

In other words, a weaker gravitational pull makes the restoring force smaller, causing the pendulum to swing more sluggishly and take more time to complete one full cycle.

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### How Gravity Fades with Altitude

Next, we need to determine how the acceleration due to gravity changes as we move away from the Earth's surface.

According to Newton's law of universal gravitation, the gravitational force is inversely proportional to the square of the distance from the center of the Earth.

At a height $h$ above the surface, the acceleration due to gravity $g(h)$ is given by:

$$g(h) = \frac{GM}{(R+h)^2}$$

We can express this in terms of the surface gravity $g = \frac{GM}{R^2}$ by factoring out $R^2$ from the denominator:

$$g(h) = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$$

This elegant formula allows us to easily calculate gravity at any altitude $h$ relative to the surface value $g$.

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### Calculating Gravity at Height $R$

Now, let's substitute our specific condition into this formula.

We are told that the second pendulum is taken to a height equal to the radius of the Earth, so we set:

$$h = R$$

Substituting this value into our gravity equation yields:

$$g_2 = \frac{g}{\left(1 + \frac{R}{R}\right)^2}$$

Since $\frac{R}{R} = 1$, the expression simplifies beautifully:

$$g_2 = \frac{g}{(1 + 1)^2} = \frac{g}{2^2} = \frac{g}{4}$$

This is a profound result!

By rising to a height of $R$ above the surface, we have doubled our distance from the Earth's center (from $R$ to $2R$).

Because of the inverse-square law, doubling the distance reduces the gravitational acceleration to exactly **one-fourth** of its surface value.

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### Finding the Ratio of Time Periods

With the local gravity at both positions known, we can now set up the ratio of their time periods, $\frac{T_2}{T_1}$.

Using our proportionality relationship:

$$\frac{T_2}{T_1} = \sqrt{\frac{g_1}{g_2}}$$

Here, $g_1$ is the gravity at the surface ($g$), and $g_2$ is the gravity at height $R$ ($\frac{g}{4}$).

Substituting these values in, we get:

$$\frac{T_2}{T_1} = \sqrt{\frac{g}{\frac{g}{4}}}$$

The surface gravity $g$ cancels out of the numerator and denominator, leaving us with:

$$\frac{T_2}{T_1} = \sqrt{4} = 2$$

Thus, the time period of the pendulum at height $R$ is exactly **twice** its time period on the surface of the Earth!

This corresponds perfectly to **option (d)**.

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### Summary and Deep Intuition

Let's take a moment to appreciate the simplicity and elegance of this result.

By moving a distance $R$ away from the surface, the gravitational field becomes $4$ times weaker.

Because the time period of a pendulum depends on the inverse square root of gravity, a $4$-fold decrease in gravity leads to a $\sqrt{4} = 2$-fold increase in the time period.

This means that if a clock regulated by this pendulum ticks once every second on Earth, it would tick once every two seconds at height $R$, running at half-speed!

This classic problem beautifully connects the geometry of gravity with the mechanics of simple harmonic motion.

Animated Solution for Physics - Gravitation: A simple pendulum has a time period $T_{1}$ when on the earth's surface and $T_{2}$ when taken to a height $R$ above the earth's surface, where $R$ is the radius of the earth. The value of $\frac{T _{2}}{T _{1}}$ is

Select Answer:

Visualized Solution

Visualizing the Two Pendulums

The Pendulum Time Period Formula

Variation of Gravity with Altitude

Substituting $h = R$

Simplifying the Gravity Expression

Setting up the Ratio $\frac{T _{2}}{T _{1}}$

Evaluating the Ratio

Conclusion and Key Takeaways

The Sigma Insight: Acceleration due to Gravity and its Variation

Introduction to the Pendulum and Gravity

The Physics of the Pendulum

How Gravity Fades with Altitude

Calculating Gravity at Height $R$

Finding the Ratio of Time Periods

Summary and Deep Intuition

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