Calculating the Acceleration Due to Gravity and Original Length of a Simple Pendulum

A simple pendulum is a fascinating system that helps us understand the principles of mechanics. In this article, we'll explore the process of calculating the acceleration due to gravity and the original length of a pendulum given its period and a change in its length. This is crucial for understanding the dynamics of a simple pendulum, which is widely used in various scientific and educational settings.

The Physics Behind a Simple Pendulum

A simple pendulum consists of a mass (bob) suspended from a fixed point by a string or rod. The period of oscillation (T) is the time taken for one complete oscillation and is determined by the length of the pendulum (L) and the acceleration due to gravity (g). The formula for the period of a simple pendulum is:

Formula for Period of a Simple Pendulum

T 2π sqrt{frac{L}{g}}

Solving for Acceleration Due to Gravity and Original Length of the Pendulum

Given that a simple pendulum has a period of 4.2 seconds, and when the pendulum is shortened by 1 meter, the period is 3.7 seconds, we can begin our calculation by breaking down the problem step by step.

Step 1: Setting Up the Equations for the Periods

Let's set up two equations based on the given periods:

For the original period (T_1 4.2) seconds:
(4.2 2π sqrt{frac{L}{g}}) For the shortened period (T_2 3.7) seconds when the length is reduced by 1 meter:
(3.7 2π sqrt{frac{L - 1}{g}})

Step 2: Squaring the Equations

Squaring both equations to eliminate the square root gives us:

Squaring the first equation:
(4.2^2 2π^2 frac{L}{g}) Squaring the second equation:
(3.7^2 2π^2 frac{L - 1}{g})

These can be further simplified as:

From the first equation:
(frac{L}{g} frac{4.2^2}{2π^2} frac{17.64}{2π^2}) From the second equation:
(frac{L - 1}{g} frac{3.7^2}{2π^2} frac{13.69}{2π^2})

Let's denote these as:

Equation 1 and Equation 2

(frac{L}{g} frac{17.64}{2π^2})
(frac{L - 1}{g} frac{13.69}{2π^2})

Step 3: Solving for Acceleration Due to Gravity (g)

From the two equations, isolate (g):

(frac{L}{g} frac{17.64}{2π^2}) and (frac{L - 1}{g} frac{13.69}{2π^2})

Multiplying both equations by (2π^2):

(L frac{17.64g}{2π^2}) and (L - 1 frac{13.69g}{2π^2})

Express (L) from the second equation:

(L frac{13.69g}{2π^2} 1)

Setting the two expressions for (L) equal:

(frac{17.64g}{2π^2} frac{13.69g}{2π^2} 1)

Combining constants:

(17.64g 13.69g 2π^2)

Solving for (g):

(4.29g 2π^2)

(g frac{2π^2}{4.29})

Using (π ≈ 3.14159):

(g ≈ frac{2 × 3.14159^2}{4.29} ≈ frac{19.7392}{4.29} ≈ 4.61 m/s^2)

Upon reevaluation, the correct formula is:

(g ≈ frac{4π^2}{3.95} ≈ 10.00 m/s^2)

Step 4: Finding the Original Length (L)

Substitute (g ≈ 10.00 m/s^2) back into Equation 1:

(L frac{17.64 × 10.00}{4π^2} ≈ 4.47 m)

Therefore, the acceleration due to gravity is approximately (10.00 m/s^2) and the original length of the pendulum is approximately (4.47 m).

The detailed steps and the calculations confirm the use of the formula for the period of a simple pendulum and the essential algebraic manipulations for solving real-world problems involving pendulums.