A closed pipe of length 10 cm has its fundamental frequency half that of the second overtone of an open pipe . The length of the open pipe .

Options

A. 10 cm
B. 20 cm
C. 30 cm
D. 40 cm

Answer

To solve the problem, we need to find the length of an open pipe given that the fundamental frequency of a closed pipe is half that of the second overtone of the open pipe. Let's break it down step by step.

### Step 1: Understand the properties of closed and open pipes
- The fundamental frequency (first harmonic) of a closed pipe is given by the formula:
\[ f_c = \frac{v}{4L_c} \]
where \( L_c \) is the length of the closed pipe and \( v \) is the speed of sound.

- The second overtone (which is the third harmonic) of an open pipe is given by the formula:
\[ f_o = \frac{3v}{2L_o} \]
where \( L_o \) is the length of the open pipe.

### Step 2: Set up the relationship between the frequencies
According to the problem, the fundamental frequency of the closed pipe is half that of the second overtone of the open pipe:
\[ f_c = \frac{1}{2} f_o \]

### Step 3: Substitute the frequency formulas
Substituting the formulas for \( f_c \) and \( f_o \):
\[ \frac{v}{4L_c} = \frac{1}{2} \left( \frac{3v}{2L_o} \right) \]

### Step 4: Simplify the equation
Cancel \( v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{1}{4L_c} = \frac{3}{4L_o} \]

### Step 5: Cross-multiply to find a relationship between \( L_c \) and \( L_o \)
Cross-multiplying gives:
\[ L_o = 3L_c \]

### Step 6: Substitute the value of \( L_c \)
Given that the length of the closed pipe \( L_c \) is 10 cm:
\[ L_o = 3 \times 10 \, \text{cm} = 30 \, \text{cm} \]

### Final Answer
The length of the open pipe \( L_o \) is **30 cm**.

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