When two sound waves having amplitude ' 3 ' and ' 5 ' units are superimposed then the ratio of maximum to minimum intensity of wave produced is

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Let the amplitudes of the two sound waves be $A_{1} = 5$ units and $A_{2} = 3$ units.

The maximum amplitude when the waves are superimposed is $A_{ma x} = A_{1} + A_{2} = 5 + 3 = 8$ units.

The minimum amplitude is $A_{min} = A_{1} - A_{2} = 5 - 3 = 2$ units.

The intensity of a wave is directly proportional to the square of its amplitude, $I \propto A^{2}$ .

The ratio of maximum to minimum intensity is given by:

$\frac{I _{ma x}}{I _{min}} = \frac{( A _{ma x} ) ^{2}}{( A _{min} ) ^{2}}$

$\frac{I _{ma x}}{I _{min}} = \frac{8 ^{2}}{2 ^{2}} = \frac{64}{4} = \frac{16}{1}$

The ratio of maximum to minimum intensity is $16 : 1$ .

Answer: 16:1

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