Two waves are simultaneously passing through a string and their equations are ${\mathrm{y}}_1={\mathrm{A}}_1\sin \mathrm{k}(\mathrm{x}-\mathrm{vt}), {\mathrm{y}}_2={\mathrm{A}}_2\sin \mathrm{k}\left(\mathrm{x}-\mathrm{vt}+{\mathrm{x}}_0\right)$.

[Given, amplitudes A₁ = 12 mm and A₂ = 5 mm, x₀ = 3.5 cm and wave number k = 6.28 cm^-1]. The amplitude of resulting wave will be .......... mm.

Given, equation of wave of first string, ${\mathrm{y}}_1={\mathrm{A}}_1\sin \mathrm{k}(\mathrm{x}-\mathrm{vt})$

Equation of wave of second string, ${\mathrm{y}}_2={\mathrm{A}}_2\sin \mathrm{k}\left(\mathrm{x}-\mathrm{vt}+{\mathrm{x}}_0\right)$

${\mathrm{A}}_1=12 \mathrm{mm}, {\mathrm{A}}_2=5 \mathrm{mm}, {\mathrm{x}}_0=3.5 \mathrm{cm}\text{ and }\mathrm{k}=6.28 {\mathrm{cm}}^{-1}$

The equation of waves after substitution of values can be expressed as

${\mathrm{y}}_1=12 \sin 6.28 (\mathrm{x}-\mathrm{vt})\text{ and }{\mathrm{y}}_2=5 \sin 6.28 (\mathrm{x}-\mathrm{vt}+3.5)$

The phase difference between two waves can be calculated as

           $\mathrm{\Delta \varphi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\mathrm{\Delta x}$

From equation of waves, the value of

$\mathrm{\Delta x}=(\mathrm{x}-\mathrm{vt}+3.5)-(\mathrm{x}-\mathrm{vt})=3.5\mathrm{cm}$

We know that, wave number is k = 2 π / λ.

Now, the phase difference is

$\mathrm{\Delta \varphi }=\mathrm{k\Delta x}=6.28\times 3.5=2\mathrm{\pi }\times 3.5=7\mathrm{\pi }$

The amplitude of resulting wave can be calculated by using the following relation,

        ${\mathrm{A}}_{\text{net }}=\sqrt{{\mathrm{A}}_1^2+{\mathrm{A}}_2^2+2{\mathrm{A}}_1{\mathrm{A}}_2\cos \mathrm{\Delta \varphi }}$

Substituting the values in above expression,

         $\begin{array}{l}{\mathrm{A}}_{\text{net }}=\sqrt{{12}^2+5^2+2\times 12\times 5\times \cos 7\mathrm{\pi }}\\ =\sqrt{144+25+120\times (-1)}\end{array}$

         ${\mathrm{A}}_{\text{net }}=\sqrt{49}=7\mathrm{mm}$

Thus, the amplitude of resultant wave is 7 mm.