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 determine the frequency \( f \) at which the man feels weightlessness at the highest point of the platform executing simple harmonic motion (SHM).

### Step 1: Understanding Weightlessness in SHM
A person feels weightlessness when the normal force acting on them becomes zero. This occurs when the acceleration of the platform at the highest point equals the acceleration due to gravity \( g \).

### Step 2: Setting Up the Equations
The displacement of the platform is given by:
\[ y = 0.5 \sin(2 \pi f t) \]
Here, \( 0.5 \) meters is the amplitude \( A \) of the motion.

### Step 3: Finding the Maximum Acceleration
The maximum acceleration \( a_{\text{max}} \) in SHM can be expressed as:
\[ a_{\text{max}} = A \omega^2 \]
where \( \omega \) is the angular frequency given by:
\[ \omega = 2 \pi f \]
Substituting \( \omega \) into the equation for maximum acceleration gives:
\[ a_{\text{max}} = A (2 \pi f)^2 \]

### Step 4: Condition for Weightlessness
At the highest point, for the man to feel weightlessness:
\[ a_{\text{max}} = g \]
Thus, we set the maximum acceleration equal to \( g \):
\[ A (2 \pi f)^2 = g \]

### Step 5: Solving for Frequency \( f \)
Substituting \( A = 0.5 \) meters into the equation:
\[ 0.5 (2 \pi f)^2 = g \]
Rearranging gives:
\[ (2 \pi f)^2 = \frac{g}{0.5} \]
\[ (2 \pi f)^2 = 2g \]
Taking the square root of both sides:
\[ 2 \pi f = \sqrt{2g} \]
Now, solving for \( f \):
\[ f = \frac{\sqrt{2g}}{2 \pi} \]

### Step 6: Substituting the Value of \( g \)
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ f = \frac{\sqrt{2 \times 9.8}}{2 \pi} \]
Calculating \( \sqrt{2 \times 9.8} \):
\[ \sqrt{19.6} \approx 4.427 \]
Thus:
\[ f \approx \frac{4.427}{2 \pi} \approx \frac{4.427}{6.283} \approx 0.705 \, \text{Hz} \]

### Final Answer
The value of \( f \) for which the man will feel weightlessness at the highest point is approximately:
\[ \boxed{0.705 \, \text{Hz}} \]