Step 1: Understanding the Concept:
When a string fixed at both ends vibrates in stationary waves, particles perform SHM.
The maximum velocity of a particle in SHM is \(v_{\text{max}} = A\omega\), where \(A\) is the amplitude of the particle and \(\omega\) is the angular frequency of the wave.
Step 2: Key Formula or Approach:
Speed of a transverse wave on a string: \(v = \sqrt{\frac{T}{\mu}}\)
Fundamental frequency: \(f_1 = \frac{v}{2L}\). For \(p\) segments, frequency is \(f_p = \frac{p \cdot v}{2L}\).
Angular frequency: \(\omega = 2\pi f\).
Step 3: Detailed Explanation:
Given: \(\mu = 0.1\text{ kg m}^{-1}\), \(L = 0.9\text{ m}\), \(T = 40\text{ N}\), \(p = 3\) segments, \(A = 0.3\text{ cm} = 3 \times 10^{-3}\text{ m}\).
First, find wave speed \(v\):
\[ v = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20\text{ m/s} \]
Now, find the vibration frequency \(f\) for 3 segments:
\[ f = \frac{3v}{2L} = \frac{3 \times 20}{2 \times 0.9} = \frac{60}{1.8} = \frac{600}{18} = \frac{100}{3}\text{ Hz} \]
Calculate angular frequency \(\omega\):
\[ \omega = 2\pi f = 2\pi \times \frac{100}{3} = \frac{200\pi}{3}\text{ rad/s} \]
Finally, calculate the maximum particle velocity:
\[ v_{\text{particle, max}} = A \cdot \omega = (3 \times 10^{-3}) \times \frac{200\pi}{3} \]
\[ v_{\text{particle, max}} = 10^{-3} \times 200\pi = 0.2\pi = \frac{2\pi}{10} = \frac{\pi}{5}\text{ m/s} \]
Step 4: Final Answer:
The maximum velocity of the particle is \(\frac{\pi}{5}\text{ m/s}\).