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: Understanding the Question:
The problem presents a coupled acoustic-mechanical resonance system. A closed organ pipe and a vibrating string are resonating with each other.
We are given the structural lengths, the tension in the string, and the speed of sound. We need to evaluate the total absolute mass of the string.

Step 2: Key Formula or Approach:
1.

Closed Pipe Fundamental Frequency ($n_p$): $$n_p = \frac{v_s}{4L_p}$$ 2.

String Harmonic Frequency ($n_s$): For a string of length $L_s$ and mass per unit length $m$ vibrating in its $p$-th harmonic: $$n_s = \frac{p}{2L_s}\sqrt{\frac{T}{m}}$$ 3.

Resonance Condition: Equate the two frequencies ($n_s = n_p$) to calculate $m$, then determine total mass $M = m \cdot L_s$.

Step 3: Detailed Explanation:
Identify the parameters given in the problem statement: Length of the closed pipe, $L_p = 0.8\text{ m}$ Length of the string, $L_s = 0.5\text{ m}$ Harmonic order of the string, $p = 2$ Tension in the string, $T = 50\text{ N}$ Speed of sound, $v_s = 320\text{ m/s}$ First, calculate the fundamental frequency of the closed organ pipe: $$n_p = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100\text{ Hz}$$ Since the string in its 2nd harmonic resonates with this frequency, set $n_s = 100\text{ Hz}$: $$100 = \frac{2}{2 \times 0.5}\sqrt{\frac{50}{m}}$$ Simplify the constant pre-factors on the right side: $$100 = \frac{1}{0.5}\sqrt{\frac{50}{m}} \implies 100 = 2\sqrt{\frac{50}{m}}$$ Divide both sides by 2: $$50 = \sqrt{\frac{50}{m}}$$ Square both sides of the equation to clear the radical wrapper: $$2500 = \frac{50}{m} \implies m = \frac{50}{2500} = \frac{1}{50} = 0.02\text{ kg/m}$$ Now, find the total absolute mass ($M$) of the string by multiplying the linear density by its length: $$M = m \times L_s = 0.02\text{ kg/m} \times 0.5\text{ m} = 0.01\text{ kg}$$ Convert kilograms to grams ($1\text{ kg} = 1000\text{ grams}$): $$M = 0.01 \times 1000 = 10\text{ grams}$$

Step 4: Final Answer:
The total mass of the string is $10\text{ grams}$, matching option (B).