Step-by-step Derivation

1. Analyze the System and Motion: The problem describes an oscillatory motion of a mass m attached to a special spring on a frictionless horizontal plane. The spring has two different spring constants: k₁ for when it is stretched (l > l₀) and k₂ for when it is compressed (l < l₀). The equilibrium position is at the natural length l₀ = 0.1 m.

   Let's define the displacement from the equilibrium position as x = l - l₀.

   • When the spring is stretched, x > 0, the restoring force is F = -k₁x. The motion is Simple Harmonic Motion (SHM) with angular frequency $ω₁ = \sqrt{k₁/m}$.
   • When the spring is compressed, x < 0, the restoring force is F = -k₂x. The motion is SHM with angular frequency $ω₂ = \sqrt{k₂/m}$.

   The total oscillation is a combination of these two SHMs. One full oscillation consists of a half-cycle in the stretched region and a half-cycle in the compressed region.

2. Calculate the Time Period of a Full Oscillation: The time period T of the full oscillation is the sum of the time taken for one half-oscillation governed by k₁ and one half-oscillation governed by k₂.

   • The time period of a full SHM with spring constant k₁ would be $T₁ = 2π / ω₁ = 2π \sqrt{m/k₁}$. The time for a half-oscillation (e.g., from one extreme position to the other) is T₁/2.
   • Similarly, the time period of a full SHM with spring constant k₂ would be $T₂ = 2π / ω₂ = 2π \sqrt{m/k₂}$. The time for a half-oscillation is T₂/2.

   The total time period T for the composite oscillation is the sum of these two half-periods: $T = \frac{T_1}{2} + \frac{T_2}{2}$ $T = \frac{1}{2} (2\pi \sqrt{\frac{m}{k_1}}) + \frac{1}{2} (2\pi \sqrt{\frac{m}{k_2}})$ $T = \pi \sqrt{\frac{m}{k_1}} + \pi \sqrt{\frac{m}{k_2}}$ $T = \pi \left( \sqrt{\frac{m}{k_1}} + \sqrt{\frac{m}{k_2}} \right)$

3. Substitute the Given Numerical Values: We are given:

   • Mass, $m = 0.1 \mathrm{~kg}$
   • Spring constant for stretching, $k₁ = 0.009 \mathrm{~N/m}$
   • Spring constant for compression, $k₂ = 0.016 \mathrm{~N/m}$

   Now, we calculate the terms inside the parenthesis: $\sqrt{\frac{m}{k_1}} = \sqrt{\frac{0.1}{0.009}} = \sqrt{\frac{100}{9}} = \frac{10}{3} \mathrm{~s}$ $\sqrt{\frac{m}{k_2}} = \sqrt{\frac{0.1}{0.016}} = \sqrt{\frac{100}{16}} = \frac{10}{4} = \frac{5}{2} \mathrm{~s}$

4. Calculate the Total Time Period T: Substitute these values back into the expression for T: $T = \pi \left( \frac{10}{3} + \frac{5}{2} \right)$ To add the fractions, we find a common denominator, which is 6: $T = \pi \left( \frac{10 \times 2}{3 \times 2} + \frac{5 \times 3}{2 \times 3} \right) = \pi \left( \frac{20}{6} + \frac{15}{6} \right)$ $T = \pi \left( \frac{20 + 15}{6} \right) = \frac{35}{6} \pi \mathrm{~s}$

5. Determine the Value of n: The problem states that the time period is $T = (n\pi) s$. Comparing this with our result: $n\pi = \frac{35}{6} \pi$ $n = \frac{35}{6}$

6. Find the Closest Integer to n: Now, we convert the fraction to a decimal: $n = \frac{35}{6} \approx 5.833...$ The integer closest to 5.833... is 6.

   (Note: The initial position $l = 0.15 \mathrm{~m}$ and natural length $l₀ = 0.1 \mathrm{~m}$ are used to define the initial state of the system, confirming it starts from an extreme position in the stretched region. However, the period of SHM is independent of the amplitude, so these values are not directly needed for the period calculation itself.)

Final Answer

The integer closest to n is 6.