Mass of Vibrating String Resonating with Closed Pipe

To find the mass of the vibrating string, we analyze the frequencies of both the closed organ pipe and the vibrating string.

1. Fundamental frequency of the closed pipe:
For a hollow pipe of length $L_p = 0.8\text{ m}$ closed at one end, the fundamental frequency $f_p$ is given by:
$f_p = \frac{v_{\text{sound}}}{4 L_p}$

Given that the speed of sound $v_{\text{sound}} = 320\text{ ms}^{-1}$:
$f_p = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100\text{ Hz}$

2. Frequency of the string:
For a string of length $L_s = 0.5\text{ m}$ fixed at both ends, the frequency of the $n^{\text{th}}$ harmonic is given by:
$f_s = n \left( \frac{v_{\text{string}}}{2 L_s} \right)$

For the second harmonic ($n = 2$):
$f_s = 2 \left( \frac{v_{\text{string}}}{2 L_s} \right) = \frac{v_{\text{string}}}{L_s}$

3. Resonance condition:
Since the string resonates with the fundamental frequency of the pipe, $f_s = f_p = 100\text{ Hz}$:
$\frac{v_{\text{string}}}{0.5} = 100$
$v_{\text{string}} = 100 \times 0.5 = 50\text{ ms}^{-1}$

4. Mass of the string:
The velocity of a transverse wave on a string is given by:
$v_{\text{string}} = \sqrt{\frac{T}{\mu}}$

where $T = 50\text{ N}$ is the tension and $\mu = \frac{m}{L_s}$ is the linear mass density of the string.

Squaring both sides:
$v_{\text{string}}^2 = \frac{T}{\mu}$
$\mu = \frac{T}{v_{\text{string}}^2} = \frac{50}{50^2} = \frac{1}{50} = 0.02\text{ kg m}^{-1}$

Now, the total mass $m$ of the string is:
$m = \mu \times L_s = 0.02\text{ kg m}^{-1} \times 0.5\text{ m} = 0.01\text{ kg} = 10\text{ grams}$

Correct Answer:
B) $10\text{ grams}$