A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion ex

Q: Why is the amplitude 2m and not 4m?

The graph shows the mean position (where U=0) is at x=2m. The extreme position is at x=4m. Amplitude is defined as the distance from the mean to the extreme position, so 4m - 2m = 2m.

Q: Can I just use the formula $U = \frac{1}{2}kx^2$?

You can only use that specific form if the mean position is at $x=0$. Here, the mean position is shifted to $x=2$, so the correct formula for potential energy at any point $x$ is $U = \frac{1}{2}k(x-2)^2$.

Q: Why do we equate the time periods of the spring and the pendulum?

The problem explicitly states that the simple pendulum has the *same period of oscillation* as the spring system. This is the crucial link that allows us to solve for gravity.

Q: Does the mass of the pendulum bob matter in this calculation?

No, the time period of a simple pendulum depends only on its length and the acceleration due to gravity. The mass of the bob cancels out in the derivation of the pendulum's time period.

Q: What if the potential energy graph was a straight line instead of a parabola?

A straight line would imply a constant force, not a restoring force proportional to displacement. Simple harmonic motion always requires a parabolic potential energy curve because the restoring force $F = -kx$ integrates to a quadratic energy function.

Question

A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?

Options:

(A) 10 m/s$^2$

(B) 5 m/s$^2$

(C) 4 m/s$^2$

(D) 9.8 m/s$^2$

Sigma Percentile · Accepted Answer

### The Interplanetary Physics Puzzle

Imagine you are an astronaut who has just landed on an unknown planet. You have two simple pieces of equipment: a spring-mass system and a simple pendulum. By observing their oscillations, you can actually calculate the acceleration due to gravity on this alien world! Let's break down exactly how this works.

### Decoding the Potential Energy Graph

First, we need to understand the spring-mass system. We are given a graph of its potential energy $U$ versus its displacement $x$.

Look closely at the curve. The lowest point of the parabola represents the **mean position** (equilibrium), where the potential energy is zero. This occurs at $x = 2	ext{ m}$. The highest points on the curve represent the **extreme positions**, where the mass momentarily stops before reversing direction. These occur at $x = 0$ and $x = 4	ext{ m}$.

The **amplitude** $A$ is the maximum displacement from the mean position.

$A = 4	ext{ m} - 2	ext{ m} = 2	ext{ m}$

At these extreme positions, the potential energy is at its maximum, which the graph shows as $U_{	ext{max}} = 10	ext{ J}$.

### Finding the Spring Constant

Now, we can use the energy formula for Simple Harmonic Motion (SHM). The maximum potential energy stored in a spring is given by:

$U_{	ext{max}} = \frac{1}{2}kA^2$

Let's substitute the values we extracted from the graph:

$10 = \frac{1}{2}k(2)^2$

$10 = 2k$

$k = 5	ext{ N/m}$

We now have the spring constant!

### The Spring's Time Period

The time period $T$ of a spring-mass system depends only on its mass $m$ and the spring constant $k$. It is completely independent of gravity. The formula is:

$T_{	ext{spring}} = 2\pi\sqrt{\frac{m}{k}}$

We are given the mass $m = 5	ext{ kg}$. Substituting our values:

$T_{	ext{spring}} = 2\pi\sqrt{\frac{5}{5}} = 2\pi	ext{ s}$

### The Pendulum Connection

Here is where the magic happens. The problem states that a simple pendulum of length $L = 4	ext{ m}$ has the **exact same period of oscillation** as our spring system on this planet.

The time period of a simple pendulum is given by:

$T_{	ext{pendulum}} = 2\pi\sqrt{\frac{L}{g}}$

Since the time periods are equal, we can equate the two expressions:

$T_{	ext{spring}} = T_{	ext{pendulum}}$

$2\pi = 2\pi\sqrt{\frac{4}{g}}$

### The Final Calculation

Let's solve for $g$. First, cancel out the $2\pi$ from both sides:

$1 = \sqrt{\frac{4}{g}}$

Square both sides to remove the square root:

$1 = \frac{4}{g}$

**$g = 4	ext{ m/s}^2$**

And there we have it! The acceleration due to gravity on this planet is $4	ext{ m/s}^2$. This elegant problem demonstrates how energy conservation and the kinematics of SHM can be linked to uncover fundamental physical constants.

Select Answer:

Visualized Solution

Problem Overview

Analyzing the Graph

Energy Equation

Calculating Spring Constant

Time Period of Spring

Spring Time Period Value

Time Period of Pendulum

Equating Time Periods

Final Calculation

The Sigma Insight: Simple Harmonic Motion (SHM)

The Interplanetary Physics Puzzle

Decoding the Potential Energy Graph

Finding the Spring Constant

The Spring's Time Period

The Pendulum Connection

The Final Calculation

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