A block with a mass of `2 kg` hangs without vibrating at the end of a spring of spring constant `500N//m`, which is attached to the ceiling of an elevator. The elevator is moving upwards with an acceleration `(g)/(3)`. At time `t = 0`, the acceleration suddenly ceases.
(a) What is the angular frequency of oscillation of the block after the acceleration ceases ?
(b) By what amount is the spring stretched during the time when the elevator is accelerating ?
(c )What is the amplitude of oscillation and initial phase angle observed by a rider in the elevator in the equation, `x = Asin (omega t + phi)` ? Take the upward direction to be positive. Take `g = 10.0 m//s^(2)`.

Explanation

Let's solve the problem step by step.

### Given Data:
- Mass of the block, \( m = 2 \, \text{kg} \)
- Spring constant, \( k = 500 \, \text{N/m} \)
- Acceleration of the elevator, \( a = \frac{g}{3} \) where \( g = 10 \, \text{m/s}^2 \)

### Part (a): Angular Frequency of Oscillation
The angular frequency \( \omega \) of a mass-spring system is given by the formula:

\[ \omega = \sqrt{\frac{k}{m}} \]

Substituting the values:

\[ \omega = \sqrt{\frac{500 \, \text{N/m}}{2 \, \text{kg}}} = \sqrt{250} \approx 15.81 \, \text{rad/s} \]

### Part (b): Amount the Spring is Stretched During Acceleration
When the elevator is accelerating upwards, the effective force on the block is modified. The net force equation can be written as:

\[ k x - mg = ma \]

Where:
- \( a = \frac{g}{3} \)

Substituting the values:

\[ k x - mg = m \left(\frac{g}{3}\right) \]

Rearranging gives:

\[ kx = mg + m\left(\frac{g}{3}\right) \]

Substituting \( m = 2 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \):

\[ kx = 2 \cdot 10 + 2 \cdot \left(\frac{10}{3}\right) = 20 + \frac{20}{3} = \frac{60}{3} + \frac{20}{3} = \frac{80}{3} \]

Now, solving for \( x \):

\[ x = \frac{80}{3k} = \frac{80}{3 \cdot 500} = \frac{80}{1500} \approx 0.0533 \, \text{m} \approx 5.33 \, \text{cm} \]

### Part (c): Amplitude of Oscillation and Initial Phase Angle
When the elevator stops accelerating, the forces acting on the block are balanced:

\[ mg = k x_0 \]

Where \( x_0 \) is the new equilibrium position. Solving for \( x_0 \):

\[ x_0 = \frac{mg}{k} = \frac{2 \cdot 10}{500} = \frac{20}{500} = 0.04 \, \text{m} = 4 \, \text{cm} \]

Now, the amplitude \( A \) of oscillation is given by:

\[ A = x - x_0 = 5.33 \, \text{cm} - 4 \, \text{cm} = 1.33 \, \text{cm} \]

To find the initial phase angle \( \phi \), we use the equation:

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

At \( t = 0 \):

\[ x = -A \Rightarrow -1.33 = 1.33 \sin(\phi) \]

This gives:

\[ \sin(\phi) = -1 \Rightarrow \phi = \frac{3\pi}{2} \, \text{radians} \]

### Summary of Results:
(a) Angular frequency \( \omega \approx 15.81 \, \text{rad/s} \)
(b) Spring stretch during acceleration \( x \approx 5.33 \, \text{cm} \)
(c) Amplitude \( A \approx 1.33 \, \text{cm} \) and initial phase angle \( \phi = \frac{3\pi}{2} \)