Question

What is the minimum distance between a node & an antinode in a standing wave?

• A. $L^2/n$
• B. $2Ln$
• C. $2L/n$
• D. $n/L$

Answer 1

Answer: (C)

2L/n (assuming a flawed question where it implicitly asks for wavelength or has a typo)

Answer 2

The minimum distance between a node and an antinode in a standing wave is always one-quarter of the wavelength ($\lambda/4$). For a string fixed at both ends or an open organ pipe, the wavelength of the $n$-th harmonic is $\lambda_n = 2L/n$. Therefore, the minimum distance between a node and an antinode would be $\frac{\lambda_n}{4} = \frac{1}{4} \left(\frac{2L}{n}\right) = \frac{L}{2n}$. For a closed organ pipe, the wavelength for the $n$-th odd harmonic is $\lambda_n = 4L/n$, and the minimum distance is $\frac{\lambda_n}{4} = \frac{1}{4} \left(\frac{4L}{n}\right) = \frac{L}{n}$. None of the provided options ($L^2/n$, $2Ln$, $2L/n$, $n/L$) match these standard results. The option $2L/n$ represents the wavelength itself for a string fixed at both ends or an open pipe, not the distance between a node and an antinode. The question as stated with the given options is flawed.