1. Fundamental frequency of an open pipe

For a tube open at both ends, the fundamental mode has $L = \frac{\lambda_A}{2}$ So, $\lambda_A = 2L$ Hence the fundamental frequency is $f_A = \frac{v}{\lambda_A} = \frac{v}{2L}$

2. Fundamental frequency of a closed pipe

For a tube closed at one end and open at the other, the fundamental mode has $L = \frac{\lambda_B}{4}$ So, $\lambda_B = 4L$ Hence the fundamental frequency is $f_B = \frac{v}{\lambda_B} = \frac{v}{4L}$

3. Find the ratio

$\frac{f_A}{f_B} = \frac{\frac{v}{2L}}{\frac{v}{4L}} = 2$ Therefore, $f_A : f_B = 2:1$

4. Option check

• A: $1:2$ ❌
• B: $1:4$ ❌
• C: $2:1$ ✅
• D: $4:1$ ❌

So the correct answer is C.