Question

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

• A. $\frac{4}{5}A$
• B. $\frac{2}{\sqrt{5}}A$
• C. $\frac{\sqrt{2}}{5}A$
• D. $\frac{1}{\sqrt{5}}A$

Answer 1

Answer: (B)

Intensity of sound wave is directly proportional to the amplitude $(A)$.
i.e., $I\propto A^2$
$\left(\frac{I_{\text{incident }}}{I_{\text{reflected }}}\right)={\left(\frac{A_{\text{incident }}}{A_{\text{reflected }}}\right)}^2$
$I_{\text{incident }}=I$
$I_{\text{reflected }}=I-20\%$ of $I$
$=I-\frac{I}{5}=\frac{4}{5}I$
$A_{\text{incident }}=A$
$\therefore$ From Eq. (i), we have
$\frac{I}{\frac{4I}{5}}=\frac{A^2}{{\left(A_{\text{reflected }}\right)}^2}$
$\Rightarrow A_{\text{reflected }}=\sqrt{\frac{4}{5}{A}^2}=\frac{2}{\sqrt{5}}A$