A hollow pipe of length `0.8m` is closed at one end. 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 ms^(-1)`, the mass of the string is

Options

A. 5 g
B. 10 g
C. 20 g
D. 40 g

Explanation

To solve the problem step by step, we can follow these calculations:

### Step 1: Calculate the fundamental frequency of the pipe
The fundamental frequency \( f \) of a closed pipe is given by the formula:
\[ f = \frac{V}{4L} \]
where:
- \( V \) is the speed of sound (320 m/s),
- \( L \) is the length of the pipe (0.8 m).

Substituting the values:
\[ f = \frac{320 \, \text{m/s}}{4 \times 0.8 \, \text{m}} = \frac{320}{3.2} = 100 \, \text{Hz} \]

### Step 2: Calculate the frequency of the vibrating string
The frequency of the string vibrating in its second harmonic is given by:
\[ f' = \frac{2V'}{2L'} \]
where:
- \( V' \) is the speed of the wave on the string,
- \( L' \) is the length of the string (0.5 m).

Since the string is vibrating in its second harmonic, we can express the speed of the wave on the string in terms of tension \( T \) and mass per unit length \( \mu \):
\[ V' = \sqrt{\frac{T}{\mu}} \]

Thus, the frequency of the string can be rewritten as:
\[ f' = \frac{2\sqrt{\frac{T}{\mu}}}{2L'} = \frac{\sqrt{\frac{T}{\mu}}}{L'} \]

### Step 3: Set the frequencies equal
Since the pipe and the string are resonating together, we set the frequencies equal:
\[ 100 \, \text{Hz} = \frac{\sqrt{\frac{T}{\mu}}}{0.5} \]

### Step 4: Solve for \( \mu \)
Rearranging the equation gives:
\[ \sqrt{\frac{T}{\mu}} = 100 \times 0.5 = 50 \]
Squaring both sides results in:
\[ \frac{T}{\mu} = 2500 \]
Thus, we can express \( \mu \) as:
\[ \mu = \frac{T}{2500} \]
Substituting \( T = 50 \, \text{N} \):
\[ \mu = \frac{50}{2500} = \frac{1}{50} \, \text{kg/m} \]

### Step 5: Convert \( \mu \) to grams per meter
To convert \( \mu \) from kg/m to grams/m:
\[ \mu = \frac{1}{50} \, \text{kg/m} = \frac{1}{50} \times 1000 \, \text{g/m} = 20 \, \text{g/m} \]

### Step 6: Calculate the mass of the string
The mass of the string \( m \) can be calculated using:
\[ m = \mu \times L' = 20 \, \text{g/m} \times 0.5 \, \text{m} = 10 \, \text{g} \]

### Final Answer
The mass of the string is \( 10 \, \text{grams} \).

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