A simple pendulum executing S.H.M. is falling freely along with the support. Then

Options

A. it does not oscillate at all
B. its periodic time increase
C. its periodic time decrease
D. None of the above

Explanation

To solve the problem of a simple pendulum executing simple harmonic motion while falling freely along with its support, we will analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the System**:
- A simple pendulum consists of a mass (bob) attached to a string or rod, which swings back and forth under the influence of gravity.
- In this scenario, the entire pendulum system is falling freely, meaning it is in a state of free fall under the influence of gravity.

2. **Frame of Reference**:
- When the pendulum is falling freely, we can analyze it from the frame of reference of the falling pendulum. In this frame, everything inside the pendulum (including the bob) experiences a condition similar to weightlessness.

3. **Applying Pseudo Force**:
- In a non-inertial frame (the frame of the falling pendulum), we need to consider a pseudo force acting on the bob. The pseudo force is equal to the mass of the bob multiplied by the acceleration of the frame, which is equal to \( mg \) (where \( m \) is the mass of the bob and \( g \) is the acceleration due to gravity).

4. **Net Effective Force**:
- The effective gravitational force acting on the bob is \( mg \) downward. However, since the entire system is falling freely, the acceleration of the bob relative to the frame is also \( g \) downward.
- Therefore, the net effective force acting on the bob can be calculated as:
\[ F_{\text{net}} = mg - mg = 0 \]
- This indicates that the bob experiences no net force.

5. **Calculating Effective Gravity**:
- The effective gravitational acceleration (\( g_{\text{effective}} \)) acting on the pendulum bob can be expressed as:
\[ g_{\text{effective}} = 0 \]

6. **Time Period of Simple Harmonic Motion**:
- The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
- Substituting \( g_{\text{effective}} = 0 \) into the formula gives:
\[ T = 2\pi \sqrt{\frac{L}{0}} \rightarrow T = \infty \]
- An infinite time period means that the pendulum does not oscillate at all.

7. **Conclusion**:
- Since the time period is infinite, the pendulum does not oscillate. Therefore, the correct option is:
- **Option 1: It does not oscillate at all.**