On getting reflected at a surface, the intensity of sound is found to be decreased by \(20 \%\). If \(A\) be…
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
\(\frac{4}{5} \mathrm{~A}\)
\(\frac{2}{\sqrt{5}} A\)
\(\frac{\sqrt{2}}{5} A\)
\(\frac{1}{\sqrt{5}} A\)
Solution
Intensity of sound wave is directly proportional to the amplitude (A).
\(\begin{aligned}
& \text{i.e., } I \propto A^2 \\
&\left(\frac{I_{\text {incident }}}{I_{\text {reflected }}}\right)=\left(\frac{A_{\text {incident }}}{\mathrm{A}_{\text {reflected }}}\right)^2 \quad \ldots (i) \\
& I_{\text {incident }}=I \\
& I_{\text {reflected }}=I-20 \% \text { of } I \\
&=I-\frac{I}{5}=\frac{4}{5} I \\
& A_{\text {incident }}=A
\end{aligned}\)
\(\therefore\) From Eq. (i), we have
\(\begin{aligned}
& \frac{I}{\frac{4 I}{5}}=\frac{A^2}{\left(A_{\text {reflected }}\right)^2} \\
\Rightarrow \quad & A_{\text {rellected }}=\sqrt{\frac{4}{5} A^2}=\frac{2}{\sqrt{5}} A
\end{aligned}\)