A simple pendulum oscillating with a small amplitude has a time period of `T = 1.0s` . A horizontal thin rod is now placed beneath the point of suspension at a distance equal to half the length of the pendulum. The string collides with the rod once in each oscillation and there is no loss of energy in such collisions. Find the new time period T of the pendulum

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Initial Conditions
The initial time period \( T \) of the simple pendulum is given as \( T = 1.0 \, \text{s} \). The time period of a simple pendulum is given by the formula:

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

where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Determine the Length of the Pendulum
From the formula, we can express the length \( L \) of the pendulum in terms of \( T \):

\[ L = \frac{g T^2}{4\pi^2} \]

### Step 3: Introduce the Rod
A horizontal thin rod is placed beneath the point of suspension at a distance equal to half the length of the pendulum. This means the rod is located at a distance \( \frac{L}{2} \) from the point of suspension.

### Step 4: Analyze the Motion After the Rod is Introduced
When the pendulum swings down, it will collide with the rod and then swing back. The effective length of the pendulum when it swings down to the rod is \( \frac{L}{2} \).

### Step 5: Calculate the New Time Period
The time period for a pendulum of length \( \frac{L}{2} \) is given by:

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

### Step 6: Combine the Time Periods
The pendulum will take time \( T' \) to swing down to the rod and the same time \( T' \) to swing back up. Therefore, the total time for one complete oscillation after the rod is introduced is:

\[ T_{\text{new}} = T' + T' = 2T' = 2\sqrt{2} T \]

### Step 7: Substitute the Initial Time Period
Since \( T = 1.0 \, \text{s} \):

\[ T_{\text{new}} = 2\sqrt{2} \cdot 1.0 \, \text{s} = 2\sqrt{2} \, \text{s} \]

### Final Answer
The new time period \( T_{\text{new}} \) of the pendulum is:

\[ T_{\text{new}} \approx 2.828 \, \text{s} \]