Question

A wave is represented by $y_1=10\cos (5x+25t)$, where $x$ and $y$ are measured in $\text{centimetres}$ and $t$ in $\text{seconds}$. A second wave for which
$y_2=20\cos (5x+25t+\pi /3)$ interferes with the first wave. Find the amplitude and phase of the resultant wave.

• A. $26.46\text{cm},0.71\text{rad}$
• B. $28.50\text{cm},1\text{rad}$
• C. $30\text{cm},0.71\text{rad}$
• D. $30.46\text{cm},0.6\text{rad}$

Answer 1

The correct option is A $26.46\text{cm},0.71\text{rad}$

Let $A_{net}$ be the amplitude and ${\varphi }_1$ be the phase of the resultant wave.
From the given wave function,
Amplitude $\left(A_1\right)$ of wave $1=10\text{cm}$
and $A_2=20\text{cm}$
Phase difference $\left(\varphi \right)=\pi /3$
From formula,
$A_{net}=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$
$=\sqrt{{10}^2+{20}^2+2\times 10\times 20\times \cos \left(\pi /3\right)}$
$=26.46\text{cm}$
From figure,
$\tan {\varphi }_1=\frac{A_2\sin \varphi }{A_1+A_2\cos \varphi }$
$\Rightarrow \tan {\varphi }_1=\frac{20\sin 60}{10+20\cos 60}=0.866$
$\Rightarrow {\varphi }_1={\tan }^{-1}\left(0.866\right)={40.89}^{\circ }$
$\Rightarrow {\varphi }_1=0.714\text{rad}$
Hence, option $\left(a\right)$ is the correct answer.