At a certain point, two waves produce pressure variations given by $\mathit{\Delta }{\mathit{p}}_{\mathbf{1}}\mathbf{=}\mathit{\Delta }{\mathit{p}}_{\mathbf{m}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\omega }\mathit{t}$and$\mathit{\Delta }{\mathit{p}}_{\mathbf{2}}\mathbf{=}\mathit{\Delta }{\mathit{p}}_{\mathbf{m}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{\omega }\mathbf{t}\mathbf{}\mathbf{-}\mathbf{\varphi }\mathbf{)}$ .At this point, what is the ratio$\mathit{\Delta }{\mathit{p}}_{\mathbf{r}}\mathbf{/}\mathit{\Delta }{\mathit{p}}_{\mathbf{m}}$ , where$\mathit{\Delta }{\mathit{p}}_{\mathbf{r}}$ is the pressure amplitude of the resultant wave, iffis (a) 0 , (b)$\mathit{\pi }\mathbf{/}\mathbf{2}$ , (c) $\mathit{\pi }\mathbf{/}\mathbf{3}$, and (d) $\mathit{\pi }\mathbf{/}\mathbf{4}$?

The pressure variations of two waves are given:

$$\begin{gathered} \Delta p_1=\Delta p_m\sin \omega t \\ \Delta p_2=\Delta p_m\sin (\omega t-\varphi ) \end{gathered}$$

The pressure variations created at a certain point by the two waves are given. Using this we can find the ratio$\mathit{\Delta }{\mathit{p}}_{\mathbf{r}}\mathbf{/}\mathit{\Delta }{\mathit{p}}_{\mathbf{m}}$ in terms of $\varphi$. Once we get the formula for ratio$\mathit{\Delta }{\mathit{p}}_{\mathbf{r}}\mathbf{/}\mathit{\Delta }{\mathit{p}}_{\mathbf{m}}$ in terms of$\varphi$ , varying the value for $\varphi$, we can find the required answer.

Formula:

The pressure amplitude of the resultant wave,

$\mathit{\Delta }{\mathit{p}}_{\mathbf{r}}\mathbf{=}\mathit{\Delta }{\mathit{p}}_{\mathbf{1}}\mathbf{+}\mathit{\Delta }{\mathit{p}}_{\mathbf{2}}$…(i)

We can use equation (i) and the given data to get the resultant pressure amplitude as given:

$\begin{array}{c}\Delta p_r=\left(\Delta p_m\sin \omega t\right)+\left(\Delta p_m\sin (\omega t-\varphi )\right)\\ \Delta p_r=\Delta p_m\times (\left(\sin \omega t\right)+\sin (\omega t-\varphi ))\\ \Delta p_r=\Delta p_m\times (2\times \cos \left(\frac{\varphi }{2}\right)\sin (\omega t-\frac{\varphi }{2}))\end{array}$

The sine function's predicate determines the sine function's amplitude$\Delta p_r$ . Thus, $\frac{\Delta p_r}{\Delta p_m}=2\times \cos \left(\frac{{\displaystyle \varphi }}{{\displaystyle 2}}\right)$ …(a)

Substituting phase, $\varphi =0$, in equation (a), we get the ratio as:

$\begin{array}{l}\frac{\Delta p_r}{\Delta p_m}=2\times \cos \left(\frac{0}{2}\right)\\ =2\times \cos \left(0\right)\\ =2\times 1\\ =2\end{array}$

Hence, the value of ratio is 2

Substituting phase, $\varphi =\pi /2$, in equation (a), we get the ratio as:

role="math" localid="1661617979917" $\begin{array}{l}\frac{\Delta p_r}{\Delta p_m}=2\times \cos \left(\frac{\frac{\pi }{2}}{2}\right)\\ =2\times \cos \left(\frac{\pi }{4}\right)\\ =2\times \left(\frac{1}{\sqrt{2}}\right)\\ =\sqrt{2}\\ =1.41\end{array}$

Hence, the value of the ratio is $1.41$.

Substituting phase,$\varphi =\pi /3$, in equation (a), we get the ratio as:

role="math" localid="1661617897406" $\begin{array}{c}\frac{\Delta p_r}{\Delta p_m}=2\times \cos \left(\frac{\frac{\pi }{3}}{2}\right)\\ =2\times \cos \left(\frac{\pi }{6}\right)\\ =2\times \left(\frac{\sqrt{3}}{2}\right)\\ =\sqrt{3}\\ =1.73\end{array}$

Hence, the value of the ratio is$1.73$ .

Substituting phase, $\varphi =\pi /4$, in equation (a), we get the ratio as:

role="math" localid="1661617723781" $\begin{array}{c}\frac{\Delta p_r}{\Delta p_m}=2\times \cos \left(\frac{\frac{\pi }{4}}{2}\right)\\ =2\times \cos \left(\frac{\pi }{8}\right)\\ =2\times \left(0.92\right)\\ =1.85\end{array}$

Hence, the value of the ratio is$1.85$ .