91Ó°ÊÓ

The pressure variations of two waves are given:

$$\begin{gathered} \mathrm{Δ}{p}_1=\mathrm{Δ}{p}_m\sin \mathrm{Ó\neg }t \\ \mathrm{Δ}{p}_2=\mathrm{Δ}{p}_m\sin (\mathrm{Ó\neg }t−\mathrm{Ï•}) \end{gathered}$$

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

Formula:

The pressure amplitude of the resultant wave,

$\mathit{Δ}{\mathit{p}}_{\mathbf{r}}\mathbf{=}\mathit{Δ}{\mathit{p}}_{\mathbf{1}}\mathbf{+}\mathit{Δ}{\mathit{p}}_{\mathbf{2}}$…(¾±)

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

$\begin{array}{c}\mathrm{Δ}{p}_r=\left(\mathrm{Δ}{p}_m\sin \mathrm{Ó\neg }t\right)+\left(\mathrm{Δ}{p}_m\sin (\mathrm{Ó\neg }t−\mathrm{Ï•})\right)\\ \mathrm{Δ}{p}_r=\mathrm{Δ}{p}_m×(\left(\sin \mathrm{Ó\neg }t\right)+\sin (\mathrm{Ó\neg }t−\mathrm{Ï•}))\\ \mathrm{Δ}{p}_r=\mathrm{Δ}{p}_m×(2×\cos \left(\frac{\mathrm{Ï•}}{2}\right)\sin (\mathrm{Ó\neg }t−\frac{\mathrm{Ï•}}{2}))\end{array}$

The sine function's predicate determines the sine function's amplitude$\mathrm{Δ}{p}_r$ . Thus, $\frac{\mathrm{Δ}{p}_r}{\mathrm{Δ}{p}_m}=2×\cos \left(\frac{{\displaystyle \mathrm{Ï•}}}{{\displaystyle 2}}\right)$ …(²¹)

Substituting phase, $\mathrm{Ï•}=0$, in equation (a), we get the ratio as:

$\begin{array}{l}\frac{\mathrm{Δ}{p}_r}{\mathrm{Δ}{p}_m}=2×\cos \left(\frac{0}{2}\right)\\ =2×\cos \left(0\right)\\ =2×1\\ =2\end{array}$

Hence, the value of ratio is 2

Substituting phase, $\mathrm{Ï•}=\mathrm{Ï€}/2$, in equation (a), we get the ratio as:

role="math" localid="1661617979917" $\begin{array}{l}\frac{\mathrm{Δ}{p}_r}{\mathrm{Δ}{p}_m}=2×\cos \left(\frac{\frac{\mathrm{π}}{2}}{2}\right)\\ =2×\cos \left(\frac{\mathrm{π}}{4}\right)\\ =2×\left(\frac{1}{\sqrt{2}}\right)\\ =\sqrt{2}\\ =1.41\end{array}$

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

Substituting phase,$\mathrm{Ï•}=\mathrm{Ï€}/3$, in equation (a), we get the ratio as:

role="math" localid="1661617897406" $\begin{array}{c}\frac{\mathrm{Δ}{p}_r}{\mathrm{Δ}{p}_m}=2×\cos \left(\frac{\frac{\mathrm{π}}{3}}{2}\right)\\ =2×\cos \left(\frac{\mathrm{π}}{6}\right)\\ =2×\left(\frac{\sqrt{3}}{2}\right)\\ =\sqrt{3}\\ =1.73\end{array}$

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

Substituting phase, $\mathrm{Ï•}=\mathrm{Ï€}/4$, in equation (a), we get the ratio as:

role="math" localid="1661617723781" $\begin{array}{c}\frac{\mathrm{Δ}{p}_r}{\mathrm{Δ}{p}_m}=2×\cos \left(\frac{\frac{\mathrm{π}}{4}}{2}\right)\\ =2×\cos \left(\frac{\mathrm{π}}{8}\right)\\ =2×\left(0.92\right)\\ =1.85\end{array}$

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