A body falling freely on a planet covers 18 m in 3 s. What is the time period of a simple pendulum of length 1 m on the planet ?

Options

A. 1.57 s
B. 3.14 s
C. 6.28 s
D. 4.5 s

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the problem
We need to find the time period of a simple pendulum of length 1 m on a planet where a body falling freely covers 18 m in 3 seconds.

### Step 2: Use the equation of motion
We can use the equation of motion for uniformly accelerated motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) = distance covered (18 m)
- \( u \) = initial velocity (0 m/s, since the body is falling freely)
- \( a \) = acceleration due to gravity (\( g' \) on this planet)
- \( t \) = time (3 s)

### Step 3: Substitute known values
Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ s = \frac{1}{2} g' t^2 \]
Substituting the known values:
\[ 18 = \frac{1}{2} g' (3^2) \]
\[ 18 = \frac{1}{2} g' (9) \]
\[ 18 = \frac{9}{2} g' \]

### Step 4: Solve for \( g' \)
To find \( g' \), we rearrange the equation:
\[ g' = \frac{18 \times 2}{9} \]
\[ g' = \frac{36}{9} \]
\[ g' = 4 \, \text{m/s}^2 \]

### Step 5: Use the formula for the time period of a simple pendulum
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
where:
- \( L \) = length of the pendulum (1 m)
- \( g' \) = acceleration due to gravity we just calculated (4 m/s²)

### Step 6: Substitute the values into the formula
Substituting \( L = 1 \, \text{m} \) and \( g' = 4 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{1}{4}} \]
\[ T = 2\pi \sqrt{0.25} \]
\[ T = 2\pi \times 0.5 \]
\[ T = \pi \]

### Step 7: Calculate the numerical value of \( T \)
Using the approximate value of \( \pi \):
\[ T \approx 3.14 \, \text{s} \]

### Final Answer
The time period of the simple pendulum of length 1 m on the planet is approximately **3.14 seconds**.

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