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
$\frac{I}{\frac{4 I}{5}}=\frac{A^{2}}{\left(A_{\text {reflected }}\right)^{2}}$
$\Rightarrow A_{\text {reflected }}=\sqrt{\frac{4}{5} A^{2}}=\frac{2}{\sqrt{5}} A$