On a frictionless horizontal plane, a bob of mass 𝑚 = 0.1 𝑘𝑔 is attached to a spring with natural length 𝑙� = 0.1 𝑚. The spring constant is 𝑘� = 0.009 𝑁𝑚�� when the length of the spring 𝑙 > 𝑙� and is 𝑘� = 0.016 𝑁𝑚�� when 𝑙 < 𝑙�. 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 𝜋) 𝑠 , then the integer closest to 𝑛 is ____.

 Alright, let's tackle this problem step by step. I'm going to break it down to understand the motion of the mass-spring system with two different spring constants depending on the spring's length.

Understanding the Problem

We have:

• A mass $m=0.1$ kg attached to a spring.

• The spring has a natural length $l_0=0.1$ m.

• The spring constant $k$ changes depending on whether the spring is stretched or compressed:

  • $k_1=0.009$ N/m when $l>l_0$ (stretched).

  • $k_2=0.016$ N/m when $l<l_0$ (compressed).

• The mass is initially released from $l=0.15$ m (which is stretched by $0.05$ m from $l_0$).

• The time period of the full oscillation is given as $T=n\pi$ seconds, and we need to find the integer closest to $n$.

Analyzing the Motion

This is a piecewise harmonic oscillator because the spring constant changes based on the displacement. The motion can be divided into two parts:

1. When the spring is stretched ($l>l_0$): governed by $k_1$.

2. When the spring is compressed ($l<l_0$): governed by $k_2$.

The mass will oscillate between the initial stretched position and some compressed position, with the spring's behavior changing at $l=l_0$.

Determining the Amplitude

At $t=0$, the mass is at $l=0.15$ m, which is $x=0.15-0.1=0.05$ m from equilibrium (stretched).

The total energy at this point is purely potential energy:

$$E=\frac{1}{2}{k}_1{x}^2=\frac{1}{2}\times 0.009\times (0.05{)}^2=\frac{1}{2}\times 0.009\times 0.0025=0.00001125\text{ J}$$

When the mass moves past $l_0$, the spring is compressed, and the spring constant changes to $k_2$. The maximum compression $x^{\mathrm{\prime }}$ can be found by equating the potential energy at maximum compression to the total energy:

$$\frac{1}{2}{k}_2{x}^{\mathrm{\prime }2}=E⟹x^{\mathrm{\prime }2}=\frac{2E}{k_2}=\frac{0.0000225}{0.016}=0.00140625⟹x^{\mathrm{\prime }}\approx 0.0375\text{ m}$$

So, the mass oscillates between $+0.05$ m (stretched) and $-0.0375$ m (compressed).

Calculating Time Period

The oscillation consists of four parts:

1. From $+0.05$ m to $0$ m (stretched, $k_1$).

2. From $0$ m to $-0.0375$ m (compressed, $k_2$).

3. From $-0.0375$ m to $0$ m (compressed, $k_2$).

4. From $0$ m to $+0.05$ m (stretched, $k_1$).

The time for each quarter period can be calculated using the formula for the time period of a simple harmonic oscillator:

$$T=2\pi \sqrt{\frac{m}{k}}$$

But since the spring constant changes, we'll calculate the time for each segment separately.

Time from $+0.05$ m to $0$ m (stretched, $k_1$):
This is a quarter period for the oscillator with $k_1$:

$$T_1=\frac{1}{4}\times 2\pi \sqrt{\frac{m}{k_1}}=\frac{\pi }{2}\sqrt{\frac{0.1}{0.009}}=\frac{\pi }{2}\sqrt{\frac{100}{9}}=\frac{\pi }{2}\times \frac{10}{3}=\frac{10\pi }{6}=\frac{5\pi }{3}$$

Wait, this seems incorrect. Actually, the time to go from maximum displacement to equilibrium is $\frac{T}{4}$, so:

$$t_1=\frac{\pi }{2}\sqrt{\frac{m}{k_1}}=\frac{\pi }{2}\sqrt{\frac{0.1}{0.009}}=\frac{\pi }{2}\times \sqrt{\frac{100}{9}}=\frac{\pi }{2}\times \frac{10}{3}=\frac{10\pi }{6}=\frac{5\pi }{3}$$

But this seems too large. Let me recheck the calculation:

$$\sqrt{\frac{0.1}{0.009}}=\sqrt{\frac{100}{9}}=\frac{10}{3}$$

So,

$$t_1=\frac{\pi }{2}\times \frac{10}{3}=\frac{10\pi }{6}=\frac{5\pi }{3}$$

This seems correct, but the value is large. Maybe the units are off. Wait, the time period $T$ is given as $n\pi$, and $\frac{5\pi }{3}$ is for a quarter period, which seems inconsistent.

Alternatively, perhaps the time from $+0.05$ to $0$ is not a full quarter period because the oscillation is not symmetric due to different $k$. Maybe we should consider the time to go from $+0.05$ to $0$ as part of the arc corresponding to $k_1$, and similarly for $k_2$.

A better approach is to consider the time spent in each region separately.

Time in stretched region ($k_1$):
The mass moves from $+0.05$ to $0$, which is like moving from amplitude to equilibrium for $k_1$:

$$t_1=\frac{T_1}{4}=\frac{\pi }{2}\sqrt{\frac{m}{k_1}}=\frac{\pi }{2}\sqrt{\frac{0.1}{0.009}}=\frac{\pi }{2}\times \frac{10}{3}=\frac{5\pi }{3}$$

Similarly, the time from $0$ to $-0.0375$ is like moving from equilibrium to amplitude for $k_2$:

$$t_2=\frac{T_2}{4}=\frac{\pi }{2}\sqrt{\frac{m}{k_2}}=\frac{\pi }{2}\sqrt{\frac{0.1}{0.016}}=\frac{\pi }{2}\times \sqrt{\frac{100}{16}}=\frac{\pi }{2}\times \frac{10}{4}=\frac{5\pi }{4}$$

Similarly, the return trip will take the same time:

• From $-0.0375$ to $0$: $t_3=t_2=\frac{5\pi }{4}$

• From $0$ to $+0.05$: $t_4=t_1=\frac{5\pi }{3}$

Total time period $T$:

$$T=t_1+t_2+t_3+t_4=\frac{5\pi }{3}+\frac{5\pi }{4}+\frac{5\pi }{4}+\frac{5\pi }{3}=2\times \frac{5\pi }{3}+2\times \frac{5\pi }{4}=\frac{10\pi }{3}+\frac{10\pi }{4}=\frac{40\pi }{12}+\frac{30\pi }{12}=\frac{70\pi }{12}=\frac{35\pi }{6}$$

Given $T=n\pi$, so:

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

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

Verifying the Calculation

Let me double-check the arithmetic:

$$\frac{10\pi }{3}+\frac{10\pi }{4}=\frac{40\pi +30\pi }{12}=\frac{70\pi }{12}=\frac{35\pi }{6}$$

So, $n=\frac{35}{6}\approx 5.833$, closest integer is $6$.

Final Answer

The integer closest to $n$ is $\boxed{{\displaystyle 6}}$.