A flat horizontal board moves up and down under SHM vertically with amplitude A. The shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is

Options

A. `2pi sqrt((g)/(A))`
B. `2pi sqrt((A)/(g))`
C. `2pi sqrt((2A)/(g))`
D. `(pi)/(2) sqrt((A)/(g))`

Answer

To solve the problem, we need to determine the shortest permissible time period of the vibration such that an object placed on the board does not lose contact with the board. Here are the step-by-step calculations:

### Step 1: Understand the Condition for Losing Contact
An object will lose contact with the board when the normal force (N) becomes zero. The forces acting on the object are its weight (Mg) acting downwards and the normal force (N) acting upwards. When the board accelerates upwards, the normal force must balance the weight of the object.

### Step 2: Apply Newton's Second Law
According to Newton's second law, the net force acting on the object is equal to the mass of the object times its acceleration. The upward acceleration of the board can be expressed in terms of angular frequency (ω) and amplitude (A) of the SHM:
\[ N + Mg = M \cdot a \]
Where \( a = \omega^2 A \) (acceleration of the board).

### Step 3: Set Normal Force to Zero
To find the condition for losing contact, we set the normal force (N) to zero:
\[ 0 + Mg = M \cdot \omega^2 A \]
This simplifies to:
\[ Mg = M \cdot \omega^2 A \]
We can cancel M from both sides (assuming M is not zero):
\[ g = \omega^2 A \]

### Step 4: Solve for Angular Frequency (ω)
Rearranging the equation gives:
\[ \omega^2 = \frac{g}{A} \]
Taking the square root:
\[ \omega = \sqrt{\frac{g}{A}} \]

### Step 5: Relate Angular Frequency to Time Period (T)
The angular frequency (ω) is related to the time period (T) by the equation:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into our expression for ω gives:
\[ \frac{2\pi}{T} = \sqrt{\frac{g}{A}} \]

### Step 6: Solve for Time Period (T)
Now, we can solve for T:
\[ T = \frac{2\pi}{\sqrt{\frac{g}{A}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{A}{g}} \]

### Final Answer
The shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is:
\[ T = 2\pi \sqrt{\frac{A}{g}} \]

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