A block is kept on a horizontal table. The table is undergoing simple farmonic motion of frequency 3Hz in a horizontal plane. The coefficient of static friction between the block and the table surface is 0.72. find the maximum amplitude of the table at which the block does not slip on the surface `g=10ms^(-1)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Frequency of the table, \( f = 3 \, \text{Hz} \)
- Coefficient of static friction, \( \mu_s = 0.72 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the angular frequency
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = 2 \pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 3 = 6 \pi \, \text{rad/s} \]

### Step 3: Set up the equation for maximum amplitude
The block will not slip if the maximum acceleration of the table is less than or equal to the maximum static frictional force. The maximum static frictional force can be expressed as:
\[ F_{\text{friction}} = \mu_s \cdot m \cdot g \]
The maximum acceleration \( a_{\text{max}} \) of the table undergoing simple harmonic motion is given by:
\[ a_{\text{max}} = \omega^2 A \]
where \( A \) is the amplitude of the motion.

### Step 4: Relate the forces
For the block to not slip:
\[ \mu_s \cdot m \cdot g \geq m \cdot a_{\text{max}} \]
This simplifies to:
\[ \mu_s \cdot g \geq a_{\text{max}} = \omega^2 A \]
Rearranging gives:
\[ A \leq \frac{\mu_s \cdot g}{\omega^2} \]

### Step 5: Substitute the values
Substituting the known values into the equation:
\[ A \leq \frac{0.72 \cdot 10}{(6 \pi)^2} \]
Calculating \( (6 \pi)^2 \):
\[ (6 \pi)^2 = 36 \pi^2 \approx 36 \times 9.87 \approx 355.44 \]
Now substituting this value back:
\[ A \leq \frac{7.2}{355.44} \approx 0.0203 \, \text{m} \]

### Step 6: Convert to centimeters
To express the amplitude in centimeters:
\[ A \approx 0.0203 \, \text{m} = 2.03 \, \text{cm} \]

### Final Answer
The maximum amplitude of the table at which the block does not slip on the surface is approximately \( 2.03 \, \text{cm} \).

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