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 motion of the plank and the mass on it, which is undergoing simple harmonic motion (SHM). The displacement of the plank is given by the equation:

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

### Step-by-Step Solution:

1. **Identify the Equation of Motion:**
The vertical displacement of the plank is given by:
\[ y = \sin(\omega t) + \sqrt{3} \cos(\omega t) \]

2. **Convert to Standard Form:**
We can express this equation in the standard form of SHM, \( y = A \sin(\omega t + \phi) \). To do this, we can find the amplitude \( A \) and the phase \( \phi \):
\[ A = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
The angle \( \phi \) can be found using:
\[ \tan(\phi) = \frac{\sqrt{3}}{1} \implies \phi = \frac{\pi}{3} \]
Thus, we can rewrite the equation as:
\[ y = 2 \sin\left(\omega t + \frac{\pi}{3}\right) \]

3. **Determine the Condition for Breaking Off:**
The mass \( m \) will just break off the plank when the normal force \( N \) becomes zero. The forces acting on the mass are:
- Gravitational force: \( mg \) (downward)
- Normal force: \( N \) (upward)

When the mass just breaks off, we have:
\[ mg = m \cdot a \]
where \( a \) is the acceleration of the plank. The acceleration in SHM is given by:
\[ a = -\omega^2 y \]
Therefore, at the point of breaking off:
\[ mg = m(-\omega^2 y) \]
Simplifying gives:
\[ g = -\omega^2 y \]

4. **Finding Minimum Value of \( \omega \):**
For the mass to just break off, we need to find the maximum value of \( y \), which is the amplitude \( A = 2 \):
\[ g = \omega^2 A \]
Substituting \( A = 2 \):
\[ g = \omega^2 \cdot 2 \implies \omega^2 = \frac{g}{2} \]
Thus, the minimum value of \( \omega \) is:
\[ \omega = \sqrt{\frac{g}{2}} \]

5. **Calculate the Time \( t \) When the Mass Breaks Off:**
We need to find the time \( t \) when the mass first breaks off. We know that this occurs when:
\[ y = 2 \implies 2 = 2 \sin\left(\omega t + \frac{\pi}{3}\right) \]
This simplifies to:
\[ 1 = \sin\left(\omega t + \frac{\pi}{3}\right) \]
The first solution for \( \sin(x) = 1 \) is:
\[ x = \frac{\pi}{2} \]
Therefore:
\[ \omega t + \frac{\pi}{3} = \frac{\pi}{2} \]
Rearranging gives:
\[ \omega t = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6} \]
Thus:
\[ t = \frac{\pi}{6\omega} \]

### Final Answers:
- The minimum value of \( \omega \) is:
\[ \omega = \sqrt{\frac{g}{2}} \]
- The time \( t \) when the mass breaks off is:
\[ t = \frac{\pi}{6\omega} \]