Question

If the amplitude of sound is doubled and the frequency is reduced to one-fourth, the intensity of sound at the same point will be

• A. increasing to a factor $2$
• B. decreasing to a factor$\frac{1}{2}$
• C. decreasing to a factor $\frac{1}{4}$
• D. unchanged

Answer 1

The correct option is C decreasing to a factor $\frac{1}{4}$

Given : $\frac{f_1}{f_2}=\frac{1}{{}^1{/}_4}=\frac{4}{1}$

$\frac{A_1}{A_2}=\frac{1}{2}$

$I=2{\pi }^2{f}^2{A}^2\rho v$

$\frac{I_1}{I_2}=\frac{2{\pi }^2{\delta }_1^2{A}_1^2\varrho v}{2{\pi }^2{f}_2^2{A}_2^2\rho v}$$=\frac{f_1^2{A}_1^2}{f_2^2{A}_2^2}$$={\left(\frac{f_1}{f_2}\right)}^2{\left(\frac{A_1}{A_2}\right)}^2$$=(4{)}^2{\left(\frac{1}{2}\right)}^2$$=16\times \frac{1}{4}$$=4$

$So,\frac{I_1}{I_2}=4$

$\Rightarrow I_2=\frac{I_1}{4}$

So, the intensity is decreased to ${\frac{1}{4}}^{th}$