Time Period of Oscillation with Asymmetric Spring Constants

To find the time period of the full oscillation of the bob, we can split the motion into two parts based on the region of oscillation:

1. Motion in the region $l > l_0$: When the spring is stretched beyond its natural length ($l > l_0$), the spring constant is $k_1 = 0.009 \text{ N m}^{-1}$. The angular frequency of oscillation in this region is: $\omega_1 = \sqrt{\frac{k_1}{m}} = \sqrt{\frac{0.009\text{ N m}^{-1}}{0.1\text{ kg}}} = \sqrt{0.09} = 0.3\text{ rad s}^{-1}$

   The bob starts from rest at $l = 0.15\text{ m}$ and moves to $l = l_0 = 0.1\text{ m}$. During a complete cycle, the bob spends half of its full SHM period of system 1 in the region $l > l_0$. The time spent in this region per oscillation is: $t_1 = \frac{T_1}{2} = \frac{\pi}{\omega_1} = \frac{\pi}{0.3} = \frac{10\pi}{3}\text{ s}$

2. Motion in the region $l < l_0$: When the spring is compressed ($l < l_0$), the spring constant is $k_2 = 0.016 \text{ N m}^{-1}$. The angular frequency of oscillation in this region is: $\omega_2 = \sqrt{\frac{k_2}{m}} = \sqrt{\frac{0.016\text{ N m}^{-1}}{0.1\text{ kg}}} = \sqrt{0.16} = 0.4\text{ rad s}^{-1}$

   Similarly, the time spent in the compressed region ($l < l_0$) during one full oscillation corresponds to half of the period of SHM of system 2: $t_2 = \frac{T_2}{2} = \frac{\pi}{\omega_2} = \frac{\pi}{0.4} = \frac{5\pi}{2}\text{ s}$

3. Total Time Period of Full Oscillation: The total time period $T$ for one complete cycle is the sum of $t_1$ and $t_2$: $T = t_1 + t_2 = \frac{10\pi}{3} + \frac{5\pi}{2} = \left( \frac{20 + 15}{6} \right) \pi = \frac{35}{6}\pi\text{ s}$

4. Finding $n$: We are given that $T = (n \pi)\text{ s}$, which gives: $n = \frac{35}{6} \approx 5.833$

   The integer closest to $5.833$ is $6$.

Final Answer: The integer closest to $n$ is 6.