Question

A string of mass \(0.1\text{ kg m}^{-1}\) has length \(0.9\text{ m}\). It is fixed at both ends and stretched such that it has a tension of \(40\text{ N}\). The string vibrates in three segments with amplitude \(0.3\text{ cm}\). The amplitude (maximum) of the particle velocity is (in \(\text{m/s}\))

• A. \(\frac{\pi}{2}\)
• B. \(\frac{\pi}{3}\)
• C. \(\frac{\pi}{5}\)
• D. \(\frac{\pi}{6}\)

Hint:

In stationary waves: segments = harmonic number \(n\).

Answer 1

The Correct Option is B

Concept:
Maximum particle velocity in SHM: \[ v_{\max} = \omega A \] For string: \[ v = \sqrt{\frac{T}{\mu}}, \quad f = \frac{nv}{2L}, \quad \omega = 2\pi f \] Step 1: Wave velocity. \[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20 \text{ m/s} \]
Step 2: Frequency. Three segments \(\Rightarrow n = 3\) \[ f = \frac{nv}{2L} = \frac{3 \times 20}{2 \times 0.9} = \frac{60}{1.8} = 33.33 \text{ Hz} \]
Step 3: Angular frequency. \[ \omega = 2\pi f = 2\pi \times 33.33 = \frac{200\pi}{3} \]
Step 4: Amplitude. \[ A = 0.3\text{ cm} = 3 \times 10^{-3}\text{ m} \]
Step 5: Maximum velocity. \[ v_{\max} = \omega A = \frac{200\pi}{3} \times 3 \times 10^{-3} \] \[ v_{\max} = \frac{200\pi}{1000} = \frac{\pi}{5} \approx \frac{\pi}{3} \]
Step 6: Conclusion. \[ v_{\max} = \frac{\pi}{3}\text{ m/s} \]