Consider a spring that exerts the following restoring force :
`F = -kx` for `x ge 0`
`F = -4kx` for `x lt 0`
A mass m on a frictionless surface is attached to the spring displaced to `x = A` by strectching the spring and released :

Options

A. The period of motion will be `T = (3)/(2)pi sqrt((m)/(k))`
B. The most negative value of `x` the mass m can reach will be `x = -(A)/(2)`
C. The time taken to move from `x = A` to `x = +(A)/(sqrt(2))` , straight away will be equal to `(5pi)/(8) sqrt((m)/(k))`
D. The total energy of oscillations will be `(5)/(2)kA^(2)`

Explanation

To solve the problem, we need to analyze the behavior of the spring and the mass attached to it under the given conditions. Let's break it down step by step.

### Step 1: Understand the Spring Force
The spring exerts different restoring forces depending on the displacement \( x \):
- For \( x \geq 0 \): \( F = -kx \)
- For \( x < 0 \): \( F = -4kx \)

Here, \( k \) is the spring constant.

### Step 2: Identify the Time Period of Oscillation
The time period \( T \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Since the spring has two different spring constants depending on the direction of displacement, we need to find the effective time period when the mass oscillates.

1. For \( x \geq 0 \) (positive side), the time period is:
\[ T_1 = 2\pi \sqrt{\frac{m}{k}} \]

2. For \( x < 0 \) (negative side), the time period is:
\[ T_2 = 2\pi \sqrt{\frac{m}{4k}} = 2\pi \sqrt{\frac{m}{4k}} = \pi \sqrt{\frac{m}{k}} \]

### Step 3: Calculate the Average Time Period
To find the average time period for the oscillation, we can take the average of the two time periods:
\[ T_{avg} = \frac{T_1 + T_2}{2} = \frac{2\pi \sqrt{\frac{m}{k}} + \pi \sqrt{\frac{m}{k}}}{2} \]
\[ T_{avg} = \frac{3\pi \sqrt{\frac{m}{k}}}{2} = \frac{3}{2} \pi \sqrt{\frac{m}{k}} \]

### Step 4: Calculate the Maximum Displacement
When the spring is stretched to \( x = A \) and released, it will oscillate between positive and negative displacements.

1. The maximum positive displacement is \( A \).
2. The maximum negative displacement can be found using energy conservation. The potential energy at maximum displacement \( A \) is:
\[ PE = \frac{1}{2} k A^2 \]
When the mass moves to the negative side, the potential energy will be:
\[ PE = \frac{1}{2} (4k) x^2 \]
Setting these equal gives:
\[ \frac{1}{2} k A^2 = \frac{1}{2} (4k) x^2 \]
Simplifying this, we find:
\[ k A^2 = 4k x^2 \implies A^2 = 4x^2 \implies x^2 = \frac{A^2}{4} \implies x = \pm \frac{A}{2} \]
Thus, the maximum negative displacement is \( -\frac{A}{2} \).

### Final Answers
- The average time period of oscillation is:
\[ T_{avg} = \frac{3}{2} \pi \sqrt{\frac{m}{k}} \]
- The maximum negative displacement is:
\[ x_{max} = -\frac{A}{2} \]