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(B) The intensity $(I)$ of a sound wave is directly proportional to the square of its amplitude $(A)$.$I \propto A^2$Therefore,the ratio of intensitie

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$\frac{2}{\sqrt{5}} A$

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(B) The intensity $(I)$ of a sound wave is directly proportional to the square of its amplitude $(A)$.$I \propto A^2$Therefore,the ratio of intensities is related to the ratio of amplitudes as:$\frac{I_{	ext{reflected}}}{I_{	ext{incident}}} = \left( \frac{A_{	ext{reflected}}}{A_{	ext{incident}}} ight)^2$Given that the intensity decreases by $20 \%$,the reflected intensity is:$I_{	ext{reflected}} = I_{	ext{incident}} - 0.20 I_{	ext{incident}} = 0.8 I_{	ext{incident}} = \frac{4}{5} I_{	ext{incident}}$Substituting this into the ratio equation:$\frac{\frac{4}{5} I_{	ext{incident}}}{I_{	ext{incident}}} = \left( \frac{A_{	ext{reflected}}}{A} ight)^2$$\frac{4}{5} = \left( \frac{A_{	ext{reflected}}}{A} ight)^2$Taking the square root on both sides:$A_{	ext{reflected}} = \sqrt{\frac{4}{5}} A = \frac{2}{\sqrt{5}} A$

On getting reflected at a surface,the intensity of sound is found to be decreased by $20 \%$. If $A$ be the amplitude of the incident sound waves,then the amplitude of reflected sound waves is

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