A man of mass `60kg` is standing on a platform executing SHM in the vertical plane. The displacement from the mean position varies as `y = 0.5sin(2pift)`. The value of `f`, for which the man will feel weightlessness at the highest point, is (`y` in metre)

Options

A. `g//4pi`
B. `4pig`
C. `sqrt(2g)/(2pi)`
D. `2pi sqrt(2g)`

Explanation

To solve the problem step by step, we need to analyze the conditions under which the man feels weightless at the highest point of the simple harmonic motion (SHM) on the platform.

### Step 1: Understanding Weightlessness in SHM
Weightlessness occurs when the net force acting on the man is zero. At the highest point in SHM, the only forces acting on him are his weight (downward) and the centripetal force required for the circular motion (upward). For him to feel weightless, these forces must balance out.

### Step 2: Setting Up the Equations
The condition for weightlessness can be expressed mathematically as:
\[ m \cdot a = m \cdot g \]
Where:
- \( m \) is the mass of the man (60 kg),
- \( a \) is the maximum acceleration of the platform,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

At the highest point, the maximum acceleration \( a \) can be expressed in terms of the angular frequency \( \omega \) and the amplitude \( A \) of the SHM:
\[ a_{\text{max}} = \omega^2 \cdot A \]

### Step 3: Relating Angular Frequency to Frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by the equation:
\[ \omega = 2 \pi f \]
Thus, we can rewrite the maximum acceleration as:
\[ a_{\text{max}} = (2 \pi f)^2 \cdot A \]

### Step 4: Substituting Values
Given that the amplitude \( A = 0.5 \, \text{m} \), we can substitute this into the equation:
\[ a_{\text{max}} = (2 \pi f)^2 \cdot 0.5 \]

### Step 5: Setting the Condition for Weightlessness
For the man to feel weightless:
\[ (2 \pi f)^2 \cdot 0.5 = g \]
Substituting \( g \approx 9.81 \, \text{m/s}^2 \):
\[ (2 \pi f)^2 \cdot 0.5 = 9.81 \]

### Step 6: Solving for Frequency \( f \)
Rearranging the equation gives:
\[ (2 \pi f)^2 = \frac{9.81}{0.5} \]
\[ (2 \pi f)^2 = 19.62 \]
Taking the square root of both sides:
\[ 2 \pi f = \sqrt{19.62} \]
\[ f = \frac{\sqrt{19.62}}{2 \pi} \]

### Step 7: Calculating the Value of \( f \)
Calculating \( \sqrt{19.62} \):
\[ \sqrt{19.62} \approx 4.43 \]
Now substituting back:
\[ f \approx \frac{4.43}{2 \pi} \]
\[ f \approx \frac{4.43}{6.2832} \]
\[ f \approx 0.706 \, \text{Hz} \]

### Final Answer
The value of \( f \) for which the man will feel weightlessness at the highest point is approximately:
\[ f \approx 0.706 \, \text{Hz} \]

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