A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest Then.

Options

A. the period of oscillation is `((2pi)/(5))`
B. `the block weighs double its actual weght, then the plank is at one of the positions of momentary rest.
C. the block weighs 1.5 times its weight on the plank halfway down
D. the block weghs its true weight on the plank when the later moves fastest.

Explanation

To solve the problem step by step, we will analyze the situation where a rectangular block is placed on a plank that is oscillating vertically in simple harmonic motion (SHM). The block loses contact with the plank when the plank is at its momentary rest position.

### Step-by-Step Solution:

1. **Understanding the System**:
- The plank oscillates vertically with an amplitude \( A = 40 \, \text{cm} = 0.4 \, \text{m} \).
- The block loses contact with the plank when the plank is at its momentary rest position.

2. **Identify Forces at the Moment of Losing Contact**:
- At the moment the block loses contact, the normal force \( N \) acting on the block becomes zero.
- The forces acting on the block are its weight \( mg \) (downward) and the net acceleration due to the plank's motion.

3. **Acceleration of the Plank**:
- The maximum acceleration \( a_{\text{max}} \) of the plank in SHM is given by:
\[ a_{\text{max}} = A \omega^2 \]
- Here, \( \omega \) is the angular frequency of the oscillation.

4. **Condition for Losing Contact**:
- At the moment of losing contact, we have:
\[ mg = m a_{\text{max}} \]
- This simplifies to:
\[ g = a_{\text{max}} = A \omega^2 \]

5. **Substituting Known Values**:
- We know \( A = 0.4 \, \text{m} \) and \( g \approx 10 \, \text{m/s}^2 \).
- Therefore:
\[ 10 = 0.4 \omega^2 \]
- Solving for \( \omega^2 \):
\[ \omega^2 = \frac{10}{0.4} = 25 \]
- Hence:
\[ \omega = 5 \, \text{rad/s} \]

6. **Finding the Period of Oscillation**:
- The period \( T \) of the oscillation is given by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting \( \omega = 5 \):
\[ T = \frac{2\pi}{5} \]

7. **Final Result**:
- The period of oscillation is:
\[ T = \frac{2\pi}{5} \, \text{s} \]