On a frictionless horizontal plane, a bob of mass $m=0.1\text{ kg}$ is attached to a spring with natural length $l_0=0.1\text{ m}$. The spring constant is $k_1=0.009{\text{ Nm}}^{-1}$ when the extension of the spring $x>0$ and is $k_2=0.016{\text{ Nm}}^{-1}$ when $x<0$. Initially the bob is released from $x=0.15\text{ m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=\pi \left(n\right)$ s, then the integer closest to n is___.

The correct answer is 6.

Step-by-Step Explanation:

1. Understanding the Motion of the Bob:
The spring system has different stiffness constants depending on whether it is extended ($x>0$) or compressed ($x<0$):
- For $x>0$, the spring constant is $k_1=0.009{\text{ N m}}^{-1}$.
- For $x<0$, the spring constant is $k_2=0.016{\text{ N m}}^{-1}$.
- The mass of the bob is $m=0.1\text{ kg}$.

2. Expression for the Total Time Period:
One complete oscillation consists of a half-cycle in the region $x>0$ with period $T_1$ and a half-cycle in the region $x<0$ with period $T_2$.

$T=\frac{T_1}{2}+\frac{T_2}{2}$

3. Calculating the Half-Period for $x>0$:

$\frac{T_1}{2}=\pi \sqrt{\frac{m}{k_1}}=\pi \sqrt{\frac{0.1}{0.009}}=\pi \sqrt{\frac{100}{9}}=\frac{10\pi }{3}\text{ s}$

4. Calculating the Half-Period for $x<0$:

$\frac{T_2}{2}=\pi \sqrt{\frac{m}{k_2}}=\pi \sqrt{\frac{0.1}{0.016}}=\pi \sqrt{\frac{100}{16}}=\frac{10\pi }{4}=\frac{5\pi }{2}\text{ s}$

5. Finding the Total Time Period $T$:

$T=\frac{10\pi }{3}+\frac{5\pi }{2}=\pi \left(\frac{10}{3}+\frac{5}{2}\right)=\pi \left(\frac{20+15}{6}\right)=\pi \left(\frac{35}{6}\right)\text{ s}$

Comparing with $T=\pi \left(n\right)$, we get:

$n=\frac{35}{6}\approx 5.833$

The integer closest to $5.833$ is 6.