An object kept on the platform a horizontal platform executing SHM, up and down with amplitude 80 cm. At what frequency will the object just lose contact with the platform?

Explanation

To solve the problem of determining the frequency at which an object just loses contact with a platform executing simple harmonic motion (SHM) with an amplitude of 80 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The object is on a platform that is oscillating vertically with a certain amplitude (A = 80 cm = 0.8 m).
- We need to find the frequency (f) at which the object will lose contact with the platform.

2. **Condition for Losing Contact**:
- The object will lose contact with the platform when the acceleration of the platform exceeds the acceleration due to gravity (g).
- The maximum acceleration (a_max) of the platform in SHM is given by the formula:
\[ a_{max} = A \omega^2 \]
where \( \omega \) is the angular frequency.

3. **Relating Angular Frequency to Frequency**:
- The angular frequency \( \omega \) is related to the frequency \( f \) by the equation:
\[ \omega = 2\pi f \]
- Substituting this into the acceleration formula gives:
\[ a_{max} = A (2\pi f)^2 = 4\pi^2 A f^2 \]

4. **Setting Up the Equation**:
- For the object to just lose contact, we set the maximum acceleration equal to the acceleration due to gravity:
\[ 4\pi^2 A f^2 = g \]

5. **Solving for Frequency**:
- Rearranging the equation to solve for \( f \):
\[ f^2 = \frac{g}{4\pi^2 A} \]
- Taking the square root gives:
\[ f = \sqrt{\frac{g}{4\pi^2 A}} \]

6. **Substituting Values**:
- Given \( g = 10 \, \text{m/s}^2 \) and \( A = 0.8 \, \text{m} \):
\[ f = \sqrt{\frac{10}{4\pi^2 \times 0.8}} \]

7. **Calculating the Frequency**:
- First, calculate \( 4\pi^2 \):
\[ 4\pi^2 \approx 39.478 \]
- Now substitute this value into the equation:
\[ f = \sqrt{\frac{10}{39.478 \times 0.8}} = \sqrt{\frac{10}{31.5824}} \approx \sqrt{0.316} \approx 0.56 \, \text{Hz} \]

### Final Answer:
The frequency at which the object just loses contact with the platform is approximately **0.56 Hz**.