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 1st overtone and resonates with the fundamental frequency of the pipe. If tension in the string is 50 N, the mass of the string is (speed of sound = 320 m/s)

• A. 25 gram
• B. 15 gram
• C. 20 gram
• D. 10 gram

Hint:

In vibrating string problems, remember that the frequency of the string is related to its length, tension, and mass. Use the appropriate formulas to connect them.

Answer 1

The Correct Option is D

Step 1: Understanding the setup.
The string is vibrating in its 1st overtone, which corresponds to the fundamental frequency of the pipe. The pipe’s fundamental frequency and the string’s frequency are related. The string's fundamental frequency can be expressed as: \[ f_{string} = \frac{v}{2L} \] where \( v \) is the speed of the wave on the string, and \( L \) is the length of the string.

Step 2: Applying the fundamental frequency relationship.
The frequency of the pipe is related to its length and the speed of sound in air. The fundamental frequency of the pipe is given by: \[ f_{pipe} = \frac{v_{sound}}{4L_{pipe}} \] where \( v_{sound} = 320 \, \text{m/s} \), and \( L_{pipe} = 0.8 \, \text{m} \). Using this, we find the frequency of the pipe.
Step 3: Solving for the mass of the string.
Using the relationship between frequency, tension, and mass per unit length of the string, we solve for the mass of the string and find it to be 10 grams.