A point object is kept in front of a plane mirror. The plane mirror is doing SHM of amplitude 2cm. The plane mirror moves along the x-axis which is normal to the mirror. The amplitude of the mirror is such that the object is always in front of the mirror. The amplitude of SHM of the image is

Options

A. `0`
B. `2cm`
C. `4cm`
D. `1cm`

Explanation

To solve the problem, we need to determine the amplitude of the image formed by a point object in front of a plane mirror that is undergoing simple harmonic motion (SHM) with a given amplitude. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
- We have a point object placed at a distance \( x \) in front of a plane mirror.
- The plane mirror is oscillating along the x-axis with an amplitude of 2 cm.

### Step 2: Determine the Position of the Image
- The image formed by a plane mirror is located at the same distance behind the mirror as the object is in front of it.
- Therefore, if the object is at position \( x \), the image will be at position \( -x \) (considering the mirror at the origin).

### Step 3: Account for the Mirror's Motion
- When the mirror moves due to SHM, it oscillates between \( -2 \) cm and \( +2 \) cm along the x-axis.
- The position of the mirror changes, affecting the position of the image.

### Step 4: Calculate the Position of the Image During SHM
- When the mirror moves to the right (positive direction), the distance from the object to the mirror decreases, and the image moves further away.
- The maximum position of the mirror is \( +2 \) cm, so the total distance from the object to the image when the mirror is at its maximum position is:
\[ \text{Total distance} = x + 2 \text{ cm} \]
- When the mirror is at its minimum position of \( -2 \) cm, the total distance becomes:
\[ \text{Total distance} = x - 2 \text{ cm} \]

### Step 5: Determine the Amplitude of the Image
- The image will oscillate between \( x + 2 \) cm and \( x - 2 \) cm.
- The total distance the image moves from its mean position (which is at \( x \)) is:
\[ \text{Amplitude of the image} = (x + 2) - (x - 2) = 4 \text{ cm} \]

### Final Answer
- Therefore, the amplitude of the image is **4 cm**.

---