Question

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

• A. 8 gram
• B. 2 gram
• C. 10 gram
• D. 4 gram

Hint:

In resonance problems, the wavelength of the vibrating string must match the wavelength of the sound wave for resonance to occur.

Answer 1

The Correct Option is C

Step 1: Understanding the resonance condition.
The pipe closed at one end has a fundamental frequency with a wavelength of \( \lambda = 4L \), where \( L \) is the length of the pipe. The string vibrating in the second harmonic must have a wavelength that matches this resonance condition. The wavelength of the string in the second harmonic is \( \lambda_s = \frac{2L}{2} \).
Step 2: Using the wave equation.
The mass per unit length of the string is \( \mu = \frac{m}{L} \), and the wave speed on the string is given by: \[ v_s = \sqrt{\frac{T}{\mu}} = 320 \, \text{m/s} \] Substituting the given values and solving for the mass, we find that the mass of the string is 10 grams.
Step 3: Conclusion.
The correct answer is (C), 10 grams.