Question

A pipe closed at one end has length $0.8\text{ m}$. At its open end, a $0.5\text{ m}$ long uniform string is vibrating in its 2nd harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50\text{ N}$ and the speed of sound is $320\text{ m/s}$, the mass of the string is

• A. $20\text{ grams}$
• B. $10\text{ grams}$
• C. $5\text{ grams}$
• D. $15\text{ grams}$

Hint:

Always distinguish carefully between the linear mass density $m$ (units of kg/m) and the total mass $M$ (units of kg or grams). Forgetting to multiply by the string's length at the end is a very common trap!
center $\text{Total Mass} = \text{Linear Density} \times \text{Length}$ center center $M = m \cdot L_s$ center

Answer 1

The Correct Option is B

Step 1: The setup.
A pipe closed at one end has length $L_p = 0.8$ m. A string of length $L_s = 0.5$ m vibrates in its 2nd harmonic and resonates with the pipe's fundamental note. The tension is $50$ N and sound speed is $320$ m/s. We want the mass of the string.
Step 2: Pipe's fundamental frequency.
A closed pipe's lowest note is \[ n_p = \frac{v_s}{4L_p} = \frac{320}{4 \times 0.8} = 100\ \text{Hz} \]
Step 3: The string's frequency.
A string of length $L_s$ and mass per length $m$ in its $p$-th harmonic gives \[ n_s = \frac{p}{2L_s}\sqrt{\frac{T}{m}} \] Here $p = 2$.
Step 4: Match the two frequencies.
Resonance means $n_s = n_p = 100$: \[ 100 = \frac{2}{2 \times 0.5}\sqrt{\frac{50}{m}} = 2\sqrt{\frac{50}{m}} \]
Step 5: Solve for mass per length.
Divide by 2: $50 = \sqrt{50/m}$. Square both sides: \[ 2500 = \frac{50}{m} \;\Rightarrow\; m = \frac{50}{2500} = 0.02\ \text{kg/m} \]
Step 6: Get the total mass.
Total mass is mass per length times length: \[ M = m \times L_s = 0.02 \times 0.5 = 0.01\ \text{kg} = 10\ \text{grams} \] This is option (2). \[ \boxed{M = 10\ \text{grams}} \]