An organ pipe of length L is open at one end | vedclass

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(A) For an organ pipe closed at one end and open at the other,the resonance condition is given by $L = (2n - 1) \frac{\lambda}{4}$,where $n = 1, 2, 3,

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$4L, 4L/3, 4L/5$

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(A) For an organ pipe closed at one end and open at the other,the resonance condition is given by $L = (2n - 1) \frac{\lambda}{4}$,where $n = 1, 2, 3, \dots$ represents the mode of vibration.Rearranging for wavelength,we get $\lambda_n = \frac{4L}{2n - 1}$.For the first three lowest frequencies (fundamental and first two overtones):For $n = 1$: $\lambda_1 = \frac{4L}{2(1) - 1} = 4L$.For $n = 2$: $\lambda_2 = \frac{4L}{2(2) - 1} = \frac{4L}{3}$.For $n = 3$: $\lambda_3 = \frac{4L}{2(3) - 1} = \frac{4L}{5}$.Thus,the wavelengths are $4L, 4L/3, 4L/5$.

An organ pipe of length $L$ is open at one end and closed at the other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

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