Question

A Pipe open at one end has length \(0.8\) m. At the open end of the tube a string \(0.5\) m long is vibrating in its first overtone and resonates with fundamental frequency of pipe. If tension in the string is \(50\) N, the mass of string is (Neglect end correction) (Speed of sound \(= 320\) m/s)

• A. \(2\) gram
• B. \(5\) gram
• C. \(10\) gram
• D. \(20\) gram

Hint:

Pipe fundamental is 100 Hz, which sets the string frequency.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A pipe open at one end has fundamental frequency \(f=\dfrac{v}{4L}\). A string fixed at both ends vibrating in its first overtone is in its second harmonic, so \(f=\dfrac{2}{2l}\sqrt{\dfrac T\mu}=\dfrac1l\sqrt{\dfrac T\mu}\).

Step 2: Pipe frequency:
\[ f=\frac{320}{4\times0.8}=100\ \text{Hz} \]

Step 3: String condition:
The string is \(l=0.5\) m long, so
\[ 100=\frac1{0.5}\sqrt{\frac{50}{\mu}}\ \Rightarrow\ \sqrt{\frac{50}\mu}=50\ \Rightarrow\ \mu=\frac{50}{2500}=0.02\ \text{kg/m} \]

Step 4: Mass of string:
\[ m=\mu l=0.02\times0.5=0.01\ \text{kg}=10\ \text{g} \]

Step 5: Choose:
Option (C).

Final Answer:
The mass of the string is 10 g. \[ \boxed{10\ \text{g}} \]