The mass and the diameter of a planet are three times the respective values for the earth. The perio

Q: The question mentions the diameter is three times, but we used the radius. Why?

The ratio of diameters is mathematically identical to the ratio of radii. If $D_p = 3D_e$, then $2R_p = 3(2R_e)$, which simplifies directly to $R_p = 3R_e$. You can safely use the same multiplier for the radius.

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

No, it doesn't! The formula for the time period of a simple pendulum is $T = 2\pi\sqrt{\frac{l}{g}}$. Notice that the mass of the bob ($m$) is completely absent from this equation. All objects fall at the same rate in a vacuum, regardless of their mass.

Q: Why did the gravity decrease even though the planet is more massive?

Gravity depends on both mass and radius. While the mass increased by a factor of 3 (which increases gravity), the radius also increased by a factor of 3. Because the radius is squared in the denominator ($3^2 = 9$), the effect of the increased distance overpowers the increased mass, resulting in a net weaker gravity.

Q: What if we took a spring pendulum to this new planet instead?

A spring pendulum's time period is given by $T = 2\pi\sqrt{\frac{m}{k}}$, where $m$ is the mass attached and $k$ is the spring constant. Since neither of these depends on gravity, the time period of a spring pendulum would remain exactly the same on the new planet!

Q: Why is $T$ inversely proportional to the square root of $g$?

Look at the formula $T = 2\pi\sqrt{\frac{l}{g}}$. If you keep $2\pi$ and $l$ constant, the only variable changing is $g$, which is in the denominator inside the square root. Thus, $T \propto \frac{1}{\sqrt{g}}$.

Question

The mass and the diameter of a planet are three times the respective values for the earth. The period of oscillation of a simple pendulum on the earth is $2	ext{ s}$. The period of oscillation of the same pendulum on the planet would be

Options:

(A) $\frac{2}{\sqrt{3}}	ext{ s}$

(B) $\frac{3}{2}	ext{ s}$

(C) $2\sqrt{3}	ext{ s}$

(D) $\frac{\sqrt{3}}{2}	ext{ s}$

Sigma Percentile · Accepted Answer

Imagine you are an astronaut carrying a simple pendulum from Earth to a newly discovered giant planet. This planet is massive—its mass and diameter are both three times that of Earth. How would your pendulum behave there? Let's embark on a journey to find out!

### Analyzing the Setup

First, let's decode the dimensions of our new planet. We are given that its mass $M_p$ is three times the mass of Earth $M_e$, so $M_p = 3M_e$.

We are also told its diameter is three times Earth's diameter. Since the radius is just half of the diameter, the ratio of the radii is exactly the same as the ratio of the diameters. Therefore, the radius of the new planet $R_p$ is three times the radius of Earth $R_e$, giving us $R_p = 3R_e$.

### The Gravity of the Situation

To understand how the pendulum swings, we need to know the gravitational pull on this new planet. According to Newton's Law of Universal Gravitation, the acceleration due to gravity $g$ on the surface of any spherical body is given by:

$$g = \frac{GM}{R^2}$$

Let's calculate the gravity on our new planet, $g_p$:

$$g_p = \frac{G(3M_e)}{(3R_e)^2}$$

Be very careful here—don't forget to square the $3$ in the denominator! Expanding the denominator gives us $9R_e^2$:

$$g_p = \frac{3GM_e}{9R_e^2} = \frac{1}{3} \left(\frac{GM_e}{R_e^2}ight)$$

Since $\frac{GM_e}{R_e^2}$ is simply the gravity on Earth ($g_e$), we find that the gravity on the new planet is surprisingly weaker:

$$g_p = \frac{1}{3} g_e$$

### The Pendulum's Rhythm

Now, let's look at our simple pendulum. The time period $T$ of a simple pendulum is determined by its length $l$ and the local gravity $g$:

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

Because we are using the exact same pendulum, its length $l$ remains constant. This means the time period is inversely proportional to the square root of gravity ($T \propto \frac{1}{\sqrt{g}}$).

Let's set up a ratio to compare the time period on the new planet ($T_p$) to the time period on Earth ($T_e$):

$$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}}$$

### Final Calculation

Substitute our finding that $g_p = \frac{g_e}{3}$ into the ratio:

$$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_e / 3}} = \sqrt{3}$$

This tells us that the pendulum will swing slower on the new planet. We are given that the time period on Earth is $T_e = 2	ext{ s}$. Multiplying this by $\sqrt{3}$, we get our final answer:

**$$T_p = 2\sqrt{3}	ext{ s}$$**

The pendulum takes $2\sqrt{3}$ seconds to complete one full oscillation on this giant, yet gravitationally weaker, planet!

Animated Solution for Physics - Oscillations: The mass and the diameter of a planet are three times the respective values for the earth. The period of oscillation of a simple pendulum on the earth is $2 s$ . The period of oscillation of the same pendulum on the planet would be

Select Answer:

Visualized Solution

Visualizing the Planets

Acceleration due to Gravity

Gravity on the New Planet

Simplifying Gravity

Time Period of Simple Pendulum

Ratio of Time Periods

Substituting Gravity Ratio

Final Calculation

The Way Forward

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Gravity of the Situation

The Pendulum's Rhythm

Final Calculation

Similar Questions

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 $T$ 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

A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period $T$ . If the mass is increased by $m$ , the time period becomes $\frac{5 T}{3}$ , then the ratio of $\frac{m}{M}$ is

If the time period of a 2 m long simple pendulum is 2 s, the acceleration due to gravity at the place, where pendulum is executing SHM is

$T_{0}$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, then the modified time period is

A ring is hung on a nail. It can oscillate without slipping or sliding (i) in its plane with a time period $T_{1}$ (ii) back and forth in a direction perpendicular to its plane, with a period $T_{2}$ . The ratio $T_{1} / T_{2}$ will be

Select Answer:

Visualized Solution

Visualizing the Planets

Acceleration due to Gravity

Gravity on the New Planet

Simplifying Gravity

Time Period of Simple Pendulum

Ratio of Time Periods

Substituting Gravity Ratio

Final Calculation

The Way Forward

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Gravity of the Situation

The Pendulum's Rhythm

Final Calculation

Similar Questions

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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

If the time period of a 2 m long simple pendulum is 2 s, the acceleration due to gravity at the place, where pendulum is executing SHM is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T0​ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to 161​ times of its initial value, then the modified time period is

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

$T_{0}$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, then the modified time period is