What is the length of a simple pendulum whose time period of oscillation is 2 second ? If this pendulum is mounted in a lift which accelerates downwards at `4 ms^(-2)` , by what factor docs its period of oscillations change from the original value?

Explanation

To solve the problem step by step, we will first find the length of the simple pendulum and then determine the factor by which the period of oscillation changes when the pendulum is in a downward-accelerating lift.

### Step 1: Find the Length of the Pendulum

The formula for the time period \( T \) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where:
- \( T \) is the time period,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Given that the time period \( T = 2 \, \text{s} \), we can rearrange the formula to solve for \( L \):

\[ 2 = 2\pi \sqrt{\frac{L}{g}} \]

Dividing both sides by \( 2\pi \):

\[ \frac{2}{2\pi} = \sqrt{\frac{L}{g}} \]

Squaring both sides:

\[ \left(\frac{1}{\pi}\right)^2 = \frac{L}{g} \]

Now, multiplying both sides by \( g \):

\[ L = g \left(\frac{1}{\pi}\right)^2 \]

Substituting \( g = 10 \, \text{m/s}^2 \):

\[ L = 10 \left(\frac{1}{\pi}\right)^2 \]

Calculating \( \pi \approx 3.14 \):

\[ L = 10 \left(\frac{1}{3.14}\right)^2 \approx 10 \left(\frac{1}{9.86}\right) \approx 1.01 \, \text{m} \]

Thus, the length of the pendulum is approximately:

\[ L \approx 1 \, \text{m} \]

### Step 2: Determine the Effective Gravity in the Lift

When the pendulum is in a lift that accelerates downwards at \( 4 \, \text{m/s}^2 \), the effective acceleration due to gravity \( g' \) is given by:

\[ g' = g - a \]

where:
- \( a = 4 \, \text{m/s}^2 \).

Substituting the values:

\[ g' = 10 - 4 = 6 \, \text{m/s}^2 \]

### Step 3: Find the New Time Period in the Lift

Using the new effective gravity, we can find the new time period \( T' \):

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

Substituting \( L \approx 1 \, \text{m} \) and \( g' = 6 \, \text{m/s}^2 \):

\[ T' = 2\pi \sqrt{\frac{1}{6}} \]

Calculating \( T' \):

\[ T' = 2\pi \cdot \frac{1}{\sqrt{6}} \approx 2\pi \cdot 0.408 \approx 2.57 \, \text{s} \]

### Step 4: Calculate the Factor of Change in Period

To find the factor by which the period changes, we can calculate:

\[ \text{Factor} = \frac{T'}{T} = \frac{2\pi \sqrt{\frac{1}{6}}}{2\pi \sqrt{\frac{1}{10}}} \]

This simplifies to:

\[ \text{Factor} = \frac{\sqrt{\frac{1}{6}}}{\sqrt{\frac{1}{10}}} = \sqrt{\frac{10}{6}} = \sqrt{\frac{5}{3}} \approx 1.29 \]

### Final Answer

1. The length of the pendulum is approximately \( 1 \, \text{m} \).
2. The factor by which the period changes when the pendulum is in the lift is approximately \( 1.29 \).