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 weight double of its weight, when the plank is at opne of the positions of momentary rest
C. the block weight `0.5` times its weight on the plank halfway up
D. the block weighs `1.5` times its weight on the plank halfway down

Explanation

To solve the problem step by step, we will analyze the situation where a rectangular block placed on a plank loses contact with the plank when the plank is at its momentary rest during vertical oscillation.

### Step 1: Understand the Problem
We have a plank that oscillates vertically with a certain amplitude, and we need to find the time period of this oscillation when the block just loses contact with the plank.

### Step 2: Identify Given Data
- Amplitude of oscillation (A) = 40 cm = 0.4 m (converted to meters for standard SI units)
- The block loses contact with the plank when the plank is at its momentary rest.

### Step 3: Analyze Forces at the Point of Losing Contact
When the block loses contact with the plank, the normal force (N) acting on the block becomes zero. At this point, the only forces acting on the block are its weight (mg) and the pseudo force due to the acceleration of the plank.

### Step 4: Write Down the Equation for Maximum Acceleration
The maximum acceleration (a_max) of the plank during simple harmonic motion is given by:
\[ a_{max} = A \cdot \omega^2 \]
where:
- A is the amplitude,
- ω is the angular frequency.

At the point of losing contact, this maximum acceleration must equal the acceleration due to gravity (g):
\[ A \cdot \omega^2 = g \]

### Step 5: Substitute Known Values
We know:
- A = 0.4 m,
- g = 10 m/s² (approximately).

Substituting these values into the equation:
\[ 0.4 \cdot \omega^2 = 10 \]

### Step 6: Solve for Angular Frequency (ω)
Rearranging the equation gives:
\[ \omega^2 = \frac{10}{0.4} = 25 \]
Taking the square root:
\[ \omega = \sqrt{25} = 5 \text{ rad/s} \]

### Step 7: Calculate the Time Period (T)
The time period (T) of oscillation is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of ω:
\[ T = \frac{2\pi}{5} \text{ seconds} \]

### Final Answer
The time period of oscillation is:
\[ T = \frac{2\pi}{5} \text{ seconds} \]

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