Physics - Sound Wave Question with Solution | TestHub

The given waves are $y_1=A\sin (\omega t-0.1x)$ and $y_2=A\sin \left(\omega t-0.1x-\frac{\varphi }{2}\right)$.

The phase difference between the two waves is $\Delta \varphi =\frac{\varphi }{2}$.

The resultant amplitude $A_R$ of two superimposed waves is given by:

$A_R=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos (\Delta \varphi )}$

Substituting $A_1=A$, $A_2=A$, and $\Delta \varphi =\frac{\varphi }{2}$:

$A_R=\sqrt{A^2+A^2+2A^2\cos \left(\frac{\varphi }{2}\right)}$

$A_R=\sqrt{2A^2\left(1+\cos \frac{\varphi }{2}\right)}$

Using the trigonometric identity $1+\cos \theta =2{\cos }^2\left(\frac{\theta }{2}\right)$:

$A_R=\sqrt{2A^2\left(2{\cos }^2\frac{\varphi }{4}\right)}$

$A_R=\sqrt{4A^2{\cos }^2\frac{\varphi }{4}}$

$A_R=2A\cos \frac{\varphi }{4}$

Answer: $2A\cos \frac{\varphi }{4}$