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

The superposition takes place between two waves of frequency $f$ and amplitude $a$. The total intensity is directly proportional to

Two waves are simultaneously passing through a string and their equations are:
${y}_{1} = {A}_{1} \sin {k}({x} - {vt}), {y}_{2} = {A}_{2} \sin {k}({x} - {vt} + {x}_{0}).$
Given amplitudes ${A}_{1} = 12 \, {mm}$ and ${A}_{2} = 5 \, {mm}$,${x}_{0} = 3.5 \, {cm}$,and wave number ${k} = 6.28 \, {cm}^{-1}$. The amplitude of the resulting wave will be $...... \, {mm}$.

Two sound waves each of intensity $I$ are superimposed. If the phase difference between the waves is $\frac{\pi}{2}$, then the intensity of the resultant wave is

The similarity between sound and light waves is:

Two waves $y_1 = A_1 \sin(\omega t - \beta_1)$ and $y_2 = A_2 \sin(\omega t - \beta_2)$ superimpose to form a resultant wave whose amplitude is: