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A closed organ pipe of length l is vibrating in $3^{r d}$ overtone with amplitude of antinode being A. Amplitude at a distance $\frac{l}{7}$ from the closed end is

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$2 A$

$\frac{A}{2}$

$\frac{A}{4}$

answer is B.

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Detailed Solution

$l = (2 n + 1) \frac{λ}{4}, R = Asinkx$

$\Rightarrow l = (2 \times 3 + 1) \frac{λ}{4} = \frac{7 λ}{4}$

$K = \frac{2 π}{λ} \Rightarrow K = \frac{2 π}{4 l} \times 7 = \frac{7 π}{2 l}$

$\Rightarrow R = Asin (\frac{7 π}{2 l} . \frac{l}{7})$

$R = Asin (\frac{π}{2})$

$∴ R = A$

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