On a frictionless horizontal plane, a bob of mass $$m = 0.1$$ kg is attached to a spring with natural length $$l_0 = 0.1$$ m. The spring constant is $$k_1 = 0.009$$ N m$$^{-1}$$ when the length of the spring $$l > l_0$$ and is $$k_2 = 0.016$$ N m$$^{-1}$$ when $$l < l_0$$. Initially the bob is released from $$l = 0.15$$ m. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is T = $$(n\pi)$$ s, then the integer closest to $$n$$ is _______.

Let the extension or compression of the spring from its natural length $$l_0$$ be denoted by $$x$$, so that
$$x = l - l_0.$$

Given data:
mass $$m = 0.1\ \text{kg}$$
spring-constant while stretched ( $$x \gt 0$$ ) : $$k_1 = 0.009\ \text{N\,m}^{-1}$$
spring-constant while compressed ( $$x \lt 0$$ ) : $$k_2 = 0.016\ \text{N\,m}^{-1}$$
initial extension $$x_0 = 0.15-0.10 = 0.05\ \text{m}$$ (released from rest).

1. Total mechanical energy of the motion
At the release point the energy is purely elastic:
$$E = \tfrac12 k_1 x_0^{2} = \tfrac12 \times 0.009 \times (0.05)^{2} = 1.125\times 10^{-5}\ \text{J}. $$

2. Maximum compression on the other side
Let the maximum compression be $$x = -x_c$$ ( $$x_c \gt 0$$ ).
At this extreme, kinetic energy is again zero, so
$$\tfrac12 k_2 x_c^{2} = E.$$
Therefore
$$x_c = x_0 \sqrt{\dfrac{k_1}{k_2}} = 0.05 \sqrt{\dfrac{0.009}{0.016}} = 0.05 \times 0.75 = 0.0375\ \text{m}.$$

The mass thus oscillates between $$x = +0.05\ \text{m}$$ (stretch) and $$x = -0.0375\ \text{m}$$ (compression).

3. Angular frequencies in the two regions
For $$x \gt 0$$ (stretch region):
$$\omega_1 = \sqrt{\dfrac{k_1}{m}} = \sqrt{\dfrac{0.009}{0.1}} = \sqrt{0.09} = 0.3\ \text{rad s}^{-1}.$$
For $$x \lt 0$$ (compression region):
$$\omega_2 = \sqrt{\dfrac{k_2}{m}} = \sqrt{\dfrac{0.016}{0.1}} = \sqrt{0.16} = 0.4\ \text{rad s}^{-1}.$$

4. Time taken in each part of the motion
In a simple harmonic motion, the time to travel from an extreme to the mean position is one-quarter of the time period of that particular SHO.
Hence
time from $$+0.05\;\text{m} \rightarrow 0$$ : $$t_1 = \dfrac{\pi}{2\omega_1}.$$
time from $$0 \rightarrow -0.0375\;\text{m}$$ : $$t_2 = \dfrac{\pi}{2\omega_2}.$$

5. Total time period
The path $$+0.05 \rightarrow 0 \rightarrow -0.0375$$ takes $$t_1 + t_2$$. The return path $$-0.0375 \rightarrow 0 \rightarrow +0.05$$ takes the same amount, so the full oscillation period is
$$T = 2\,(t_1 + t_2) = 2\!\left( \dfrac{\pi}{2\omega_1} + \dfrac{\pi}{2\omega_2} \right) = \pi\!\left( \dfrac{1}{\omega_1} + \dfrac{1}{\omega_2} \right).$$

Substituting $$\omega_1 = 0.3,\; \omega_2 = 0.4$$:
$$T = \pi \left( \dfrac{1}{0.3} + \dfrac{1}{0.4} \right) = \pi (3.333\dots + 2.5) = 5.833\dots\,\pi\ \text{s}.$$

6. Writing in the required form
The period is given as $$T = (n\pi)\ \text{s}$$, hence $$n = 5.833\dots$$.
The integer closest to $$n$$ is $$6$$.

Therefore, the required integer is 6.