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:

Find the wave speed and the frequency of the three-segment mode, then the maximum particle velocity is amplitude times angular frequency.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Wave speed \(v=\sqrt{T/\mu}=\sqrt{\frac{40}{0.1}}=20\text{ m/s}\).

Step 2: Wavelength
Three segments between fixed ends: \(L=\frac{3\lambda}{2}\), so \(\lambda=\frac{2\times0.9}{3}=0.6\text{ m}\).

Step 3: Frequency
\(f=\frac v\lambda=\frac{20}{0.6}=\frac{100}{3}\text{ Hz}\), so \(\omega=2\pi f=\frac{200\pi}{3}\text{ rad/s}\).

Step 4: Maximum particle velocity
\(v_{max}=A\omega=0.003\times\frac{200\pi}{3}=0.2\pi=\frac{\pi}{5}\text{ m/s}\).

Step 5: Compare the options
The value is \(\frac{\pi}{5}\text{ m/s}\), the third option.

Final Answer:
The amplitude of the particle velocity is \(\frac{\pi}{5}\text{ m/s}\). \[ \boxed{\text{(C) }\dfrac{\pi}{5}\text{ m/s}} \]