An object of mass `m` is attached to a spring. The restroing force of the spring is `F = - lambdax^(3)`, where `x` is the displacement. The oscillation period depends on the mass, `l`mabd and oscillation amplitude. Suppose the object is initially at rest. If the initial displacement is `D` then its period is `tau`. If the initial displacement is `2D`, find the period.

Options

A. `8tau`
B. `2tau`
C. `tau`
D. `tau//2`

Explanation

To solve the problem, we need to find the relationship between the period of oscillation and the amplitude of the oscillation for the given restoring force \( F = -\lambda x^3 \).

### Step-by-Step Solution

1. **Understanding the Force**: The restoring force of the spring is given by \( F = -\lambda x^3 \). This indicates that the force is non-linear and depends on the cube of the displacement \( x \).

2. **Identifying the Period Dependence**: The period \( T \) of oscillation for a system can generally be expressed as a function of mass \( m \), spring constant \( \lambda \), and amplitude \( A \). We can express this as:
\[ T = k \cdot m^a \cdot \lambda^b \cdot A^c \]
where \( k \) is a constant, and \( a, b, c \) are the powers to which each variable is raised.

3. **Finding the Dimensions**: We need to analyze the dimensions of each variable:
- Mass \( m \) has the dimension \( [M] \).
- The spring constant \( \lambda \) has dimensions derived from the force equation. Since \( F = m \cdot a \) and \( a = \frac{d^2x}{dt^2} \), we can express \( \lambda \) as having dimensions \( [M][L^{-2}] \) when considering the force \( F = -\lambda x^3 \).

4. **Setting Up the Equation**: From the dimensional analysis, we can write:
\[ [T] = [M^a][\lambda^b][A^c] \implies [T] = [M^a][M^b][L^{-2b}][L^c] \]
This simplifies to:
\[ [T] = [M^{a+b}][L^{c-2b}] \]

5. **Equating Dimensions**: Since the dimension of time \( [T] \) is \( [L^0][M^0][T^1] \), we can set up the following equations:
- For mass: \( a + b = 0 \)
- For length: \( c - 2b = 1 \)

6. **Solving the Equations**:
- From \( a + b = 0 \), we have \( a = -b \).
- Substituting \( a = -b \) into \( c - 2b = 1 \):
\[ c - 2(-a) = 1 \implies c + 2a = 1 \]
- We can express \( c \) in terms of \( a \):
\[ c = 1 - 2a \]

7. **Finding the Relationship**: Now, we know that the period \( T \) depends on the amplitude \( A \). If we assume \( A = D \) gives period \( \tau \), then:
\[ \tau \propto D^{c} \implies \tau = k \cdot m^a \cdot \lambda^b \cdot D^{c} \]
When \( A = 2D \):
\[ T' = k \cdot m^a \cdot \lambda^b \cdot (2D)^{c} = k \cdot m^a \cdot \lambda^b \cdot 2^{c} \cdot D^{c} \]
Thus:
\[ T' = 2^{c} \cdot \tau \]

8. **Substituting for \( c \)**: Since \( c = 1 - 2a \), we can substitute this into the equation:
\[ T' = 2^{(1 - 2a)} \cdot \tau \]

9. **Conclusion**: The period when the initial displacement is \( 2D \) can be expressed as:
\[ T' = 2^{(1 - 2a)} \cdot \tau \]

### Final Result
The period when the initial displacement is \( 2D \) is \( T' = 2^{(1 - 2a)} \cdot \tau \).