Question

The fundamental frequency of a closed organ pipe of length $20\,cm$ is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

• A. 120 cm
• B. 140 cm
• C. 80 cm
• D. 100 cm

Answer 1

The Correct Option is A

The problem involves comparing the fundamental frequency of a closed organ pipe with the second overtone of an open organ pipe. Let's solve this step by step.

Step-by-Step Solution:

1. The fundamental frequency of a closed organ pipe of length L_c = 20 \, \text{cm} (0.2 m) can be calculated using the formula: f_c = \dfrac{v}{4L_c}, where v is the speed of sound in air (approximately 340 \, \text{m/s}).
2. The frequency of the nth overtone for an open pipe is given by: f_n = \dfrac{nv}{2L_o}, where L_o is the length of the open pipe and n is the overtone number.
3. The second overtone of an open pipe (which corresponds to the first overtone) is f_2 = \dfrac{3v}{2L_o}.
4. According to the given problem, the fundamental frequency of the closed pipe (f_c) is equal to the second overtone of the open pipe (f_2): \dfrac{v}{4L_c} = \dfrac{3v}{2L_o}.
5. We can simplify this equation by cancelling v and cross-multiplying: 2L_o = 12L_c.
6. Solving for L_o, we get: L_o = 6L_c = 6 \times 0.2 \, \text{m} = 1.2 \, \text{m} = 120 \, \text{cm}.

Therefore, the length of the organ pipe open at both ends is 120 cm, matching the given correct option.