91Ó°ÊÓ

An organ pipe is a cylindrical tube in which a standing wave is produced due to the vibrations of air particles. There are two types of organ pipes - open at both ends and closed at one end.

For an organ pipe open at both ends, resonance occurs when the length can accommodate an integer number of half-wavelengths. For the nth mode of vibration, the condition for resonance is given by: \(L_{open} = n \cdot \dfrac{\lambda_{open}}{2}\), where \(L_{open}\) is the length of the open pipe, \(\lambda_{open}\) is the wavelength, and n is an integer (called the harmonic number).

For an organ pipe closed at one end, resonance occurs when the length can accommodate an odd number of quarter-wavelengths. For the nxt mode of vibration, the condition for resonance is given by: \(L_{closed} = n \cdot \dfrac{\lambda_{closed}}{4}\), where \(L_{closed}\) is the length of the closed pipe, \(\lambda_{closed}\) is the wavelength, and nx is an odd integer.

Now, let's compare the resonance conditions of open and closed pipes to determine the ratio of their lengths: \(L_{open} = n \cdot \dfrac{\lambda_{open}}{2}\) \(L_{closed} = n_x \cdot \dfrac{\lambda_{closed}}{4}\) We need to find the smallest whole numbers n and nx such that the lengths resonate with each other. This happens when we choose \(n = 2\) for the open pipe (second harmonic) and \(n_x = 1\) for the closed pipe (first harmonic). The resulting length resonance condition becomes: \(L_{open} = 2 \cdot \dfrac{\lambda_{open}}{2}\) \(L_{closed} = 1 \cdot \dfrac{\lambda_{closed}}{4}\) Now, since both the pipes resonate together, their fundamental frequencies also need to match, which means \(v_{open} = v_{closed}\). The speed of sound (v) in both pipes is the same; hence we can write: \(\dfrac{\lambda_{open}}{T_{open}} = \dfrac{\lambda_{closed}}{T_{closed}}\) Knowing that \(T_{open} = 2 \cdot T_{closed}\), we can write: \(\dfrac{\lambda_{open}}{2 \cdot T_{closed}} = \dfrac{\lambda_{closed}}{T_{closed}}\) Simplifying, we get: \(\lambda_{open} = 2 \cdot \lambda_{closed}\) Now substituting the wavelengths expressions from the resonance conditions: \(2 \cdot L_{closed} = L_{open}\) Finally, we get the ratio between the lengths of organ pipes: \(\dfrac{L_{open}}{L_{closed}} = \dfrac{2 \cdot L_{closed}}{L_{closed}} = 2\) Thus, the correct answer is (C) 2: 1.