A plank with a bar placed on it performs horizontal harmonic oscillations with amplitude `a=10cm`. Find the coefficient of friction between the bar and the plank if the former starts sliding along the plank when the amplitude of oscillation of the plank becomes less than `T=1.0 s`.

Explanation

To solve the problem, we need to find the coefficient of friction (μ) between the bar and the plank, given that the bar starts sliding when the amplitude of oscillation of the plank becomes less than T = 1.0 s.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- The bar is placed on a plank that performs horizontal harmonic oscillations with an amplitude \( a = 10 \, \text{cm} = 0.1 \, \text{m} \).
- The bar will start sliding when the maximum acceleration of the plank exceeds the maximum static friction force acting on the bar.

2. **Maximum Acceleration in SHM:**
- The maximum acceleration \( a_{\text{max}} \) in Simple Harmonic Motion (SHM) is given by the formula:
\[ a_{\text{max}} = a \omega^2 \]
- Where \( a \) is the amplitude and \( \omega \) is the angular frequency.

3. **Finding Angular Frequency (ω):**
- The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
- Given \( T = 1.0 \, \text{s} \):
\[ \omega = \frac{2\pi}{1} = 2\pi \, \text{rad/s} \]

4. **Calculating Maximum Acceleration:**
- Substitute \( a = 0.1 \, \text{m} \) and \( \omega = 2\pi \):
\[ a_{\text{max}} = 0.1 \times (2\pi)^2 \]
- Calculate \( (2\pi)^2 \):
\[ (2\pi)^2 = 4\pi^2 \]
- Therefore:
\[ a_{\text{max}} = 0.1 \times 4\pi^2 \]

5. **Friction Force:**
- The maximum static friction force \( F_s \) is given by:
\[ F_s = \mu mg \]
- Where \( m \) is the mass of the bar and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

6. **Setting Up the Equation:**
- For the bar to start sliding, the maximum static friction must equal the maximum acceleration force:
\[ \mu mg = ma_{\text{max}} \]
- Canceling \( m \) from both sides gives:
\[ \mu g = a_{\text{max}} \]

7. **Substituting Values:**
- Substitute \( g = 10 \, \text{m/s}^2 \) and \( a_{\text{max}} = 0.1 \times 4\pi^2 \):
\[ \mu \times 10 = 0.1 \times 4\pi^2 \]
- Rearranging gives:
\[ \mu = \frac{0.1 \times 4\pi^2}{10} \]
- Simplifying:
\[ \mu = 0.01 \times 4\pi^2 \]

8. **Approximating π²:**
- Using \( \pi^2 \approx 10 \):
\[ \mu = 0.01 \times 4 \times 10 = 0.4 \]

### Final Answer:
The coefficient of friction \( \mu \) between the bar and the plank is **0.4**.