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A source emitting sound of frequency $90 \mathrm{~Hz}$ is Placed in front of a wall at a distance $8 \mathrm{~m}$ from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and the detector for which the detector detects a maximum of sound speed in air $340 \mathrm{~m} / \mathrm{s}$ .
Option: 1
67.84

Option: 2
66.84

Option: 3
65.84

Option: 4
68.84

Answers (1)

best_answer

$\Delta x=$ Path difference

$=n \lambda[n=1] \\$

$=\lambda \\$

$2 \sqrt{x^2+64}+\frac{\lambda}{2}-2 x=\lambda$

$\lambda= \frac{v}{f}=\frac{340}{90} \\$ $\lambda= \frac{V}{f}=\frac{340}{90} \\$

$\lambda= \frac{34}{9} \\$

$2 \sqrt{x^2+64}+\frac{34}{18}-2 x=\frac{34}{9} \\$

$2 \sqrt{x^2+64}-2 x=\frac{34}{9}-\frac{34}{18}$

$2 \sqrt{x^2+64}-2 x=\frac{17}{9} \\$

$2 \sqrt{x^2+64}=\frac{17}{9}+2 x \\$

$18 \sqrt{x^2+64}=18 x+17 \\$

$324\left(x^2+64\right)=324 x^2+289+612 x \\$

$324 x^2+20736=324 x^2+289+612 x \\$

$20736-289=612 x$

$20447=612 x \\$

$x=\frac{20447}{612} \\$

$x=33.42 \\$

$x=2 x \\$

$x=2 \times 33.42 \\$

$x=66.84$

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