A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude A. What will be the frequency of oscillation, the block will just start to slip? Coefficient of friction`=mu`.

Options

A. `(1)/(2pi)sqrt((mug)/(A))`
B. `(1)/(4pi)sqrt((mug)/(A))`
C. `2pisqrt((A)/(mug))`
D. `4pisqrt((A)/(mug))`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and relate them to the frequency of oscillation of the table.

### Step 1: Understand the forces acting on the block
The block is resting on a table that is executing simple harmonic motion (SHM). The key forces acting on the block are:
- The gravitational force (weight) acting downwards, which is \( mg \).
- The normal force \( N \) acting upwards, which balances the weight of the block in the vertical direction.
- The frictional force that prevents the block from slipping, which is given by \( F_f = \mu N \).

Since there are no vertical accelerations, we have:
\[ N = mg \]

### Step 2: Determine the maximum acceleration of the table
The table is executing SHM with an amplitude \( A \). The maximum acceleration \( a_{max} \) of the table can be expressed as:
\[ a_{max} = \omega^2 A \]
where \( \omega \) is the angular frequency of the SHM.

### Step 3: Relate the maximum acceleration to the frictional force
For the block to just start slipping, the maximum frictional force must equal the net force acting on the block due to the acceleration of the table. The maximum frictional force is:
\[ F_f = \mu N = \mu mg \]

Setting the maximum frictional force equal to the force due to the maximum acceleration:
\[ \mu mg = m a_{max} \]
Substituting for \( a_{max} \):
\[ \mu mg = m \omega^2 A \]

### Step 4: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = \omega^2 A \]

### Step 5: Solve for the angular frequency \( \omega \)
Rearranging the equation gives:
\[ \omega^2 = \frac{\mu g}{A} \]
Taking the square root of both sides:
\[ \omega = \sqrt{\frac{\mu g}{A}} \]

### Step 6: Convert angular frequency to frequency
The frequency \( f \) is related to angular frequency \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
Substituting for \( \omega \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{\mu g}{A}} \]

### Final Answer
Thus, the frequency of oscillation at which the block will just start to slip is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{\mu g}{A}} \]