Vertical displacement of a plank with a body of mass m on it is varying according to the law `y=sinomegat+sqrt(3)cosomegat`. The minimum value of `omega` for which the mass just breaks off the plank and the moment it occurs first time after t=0, are given by (y is positive towards vertically upwards).

Options

A. `sqrt(g/2),sqrt(2/6),(pi)/(sqrt(g))`
B. `g/sqrt(2),2/3sqrt(pi)/g)`
C. `sqrt(g/2),(pi)/3)sqrt(2/g)`
D. `sqrt(2g),sqrt((2pi)/(3g))`

Answer

To solve the problem, we need to analyze the vertical displacement of the plank with the mass on it, given by the equation:

\[ y = \sin(\omega t) + \sqrt{3} \cos(\omega t) \]

### Step 1: Rewrite the equation in a standard form
We can express the equation in a more recognizable form by combining the sine and cosine terms. We can use the identity for sine of a sum:

\[ y = A \sin(\omega t + \phi) \]

where \( A \) is the amplitude and \( \phi \) is the phase angle. To find \( A \) and \( \phi \), we calculate:

- \( A = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \)
- \( \tan(\phi) = \frac{\text{coefficient of } \sin}{\text{coefficient of } \cos} = \frac{1}{\sqrt{3}} \Rightarrow \phi = 30^\circ \) or \( \frac{\pi}{6} \) radians.

Thus, we can rewrite the equation as:

\[ y = 2 \sin\left(\omega t + \frac{\pi}{6}\right) \]

### Step 2: Determine the conditions for the mass to lose contact with the plank
The mass will lose contact with the plank when the normal force \( N \) becomes zero. The forces acting on the mass are:

- Weight \( mg \) acting downwards.
- Normal force \( N \) acting upwards.

Using Newton's second law, we have:

\[ N - mg = ma \]

where \( a \) is the acceleration of the mass. When the mass just loses contact, \( N = 0 \), thus:

\[ -mg = ma \]
\[ a = -g \]

### Step 3: Relate acceleration to the motion of the plank
The acceleration \( a \) of the mass is given by the second derivative of the displacement \( y \):

\[ a = \frac{d^2y}{dt^2} = -\omega^2 y \]

At the point of losing contact, we have:

\[ -\omega^2 y = -g \]

Substituting \( y \) at its maximum value (which is the amplitude \( A = 2 \)):

\[ \omega^2 \cdot 2 = g \]
\[ \omega^2 = \frac{g}{2} \]
\[ \omega = \sqrt{\frac{g}{2}} \]

### Step 4: Calculate the minimum value of \( \omega \)
Using \( g \approx 9.8 \, \text{m/s}^2 \):

\[ \omega = \sqrt{\frac{9.8}{2}} = \sqrt{4.9} \approx 2.21 \, \text{rad/s} \]

### Step 5: Determine the time when the mass first loses contact
To find the time \( t \) when the mass first loses contact, we set \( y \) equal to the maximum amplitude:

\[ y = 2 \sin\left(\omega t + \frac{\pi}{6}\right) = 2 \]

This occurs when:

\[ \sin\left(\omega t + \frac{\pi}{6}\right) = 1 \]

The first occurrence of this happens at:

\[ \omega t + \frac{\pi}{6} = \frac{\pi}{2} \]
\[ \omega t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

Thus:

\[ t = \frac{\pi/3}{\omega} = \frac{\pi/3}{\sqrt{4.9}} \]

Calculating \( t \):

\[ t \approx \frac{1.047}{2.21} \approx 0.474 \, \text{s} \]

### Final Answers
- The minimum value of \( \omega \) is approximately \( 2.21 \, \text{rad/s} \).
- The time \( t \) when the mass first loses contact is approximately \( 0.474 \, \text{s} \).