91Ó°ÊÓ

• Wavelength of the sound is, $\mathrm{λ}=2\text{m}$.

• Separation between the source,$d=1\text{6m}$.

Sound waves are sinusoidal waves, which consist of wavelength and frequency. Calculate the phase difference between the two waves.

Formula is as follow:

$f=\frac{2\mathrm{π}}{\mathrm{λ}}\text{Δ}L$

Here, L is length, $\mathrm{λ}$ is wavelength and f is frequency.

Point P is infinitely far from the source. Therefore, the small distance d in between the two sources is of no consequence. Thus, there is no effect of distance on the phase difference. Therefore, there is no phase difference between the two waves.

$\begin{array}{c}f=\frac{2\mathrm{π}}{\mathrm{λ}}\text{Δ}L\\ =\frac{2\mathrm{π}}{\mathrm{λ}}×0\\ =0\end{array}$

Hence, the phase difference between $s_1$and $s_2$is zero.

If two sources oscillate inthephase, they produce constructive interference. In this problem, source$s_1$and source$s_2$are inthatphase, so they produce constructive interference.

Hence, type of interference produced by $s_1$and$s_2$is constructive.

Now, point P is moving towards source. Supposethedistance between source$s_1$and point P is x. By using Pythagorean theorem, findthedistance between source$s_2$and point p as,

$q=\sqrt{d^2+x^2}$$x>0,$

So, that path difference between two waves is,

$\text{Δ}L=\sqrt{d^2+x^2}−x$

Where, i.e., the path of the two sources is different. So, the phase difference has non-zero value, and it is clearly increasing.

Hence,the phase difference betweenthewaves increases.

From the above explanation, write the phase difference as,

$\begin{array}{l}f=\frac{\text{Δ}L}{\mathrm{λ}}\\ f=\frac{\sqrt{d^2+x^2}−x}{\mathrm{λ}}\end{array}$

In terms of $\mathrm{λ},$the phase difference is identical to the path length difference,

$\begin{array}{l}\text{Δ}L=\frac{\mathrm{λ}}{2}\\ \frac{\text{Δ}L}{\mathrm{λ}}=\frac{\sqrt{d^2+x^2}−x}{\mathrm{λ}}\end{array}$

On solving the above equation,

$$\begin{gathered} \text{Δ}L=\sqrt{d^2+x^2}−x \\ \frac{\mathrm{λ}}{2}=\sqrt{d^2+x^2}−x \\ x+\frac{\mathrm{λ}}{2}=\sqrt{d^2+x^2} \end{gathered}$$

Squaring both sides,

$\begin{array}{c}{x}^2+\frac{{\mathrm{λ}}^2}{4}+x\mathrm{λ}=d^2+x^2\\ \frac{{\mathrm{λ}}^2}{4}+x\mathrm{λ}=d^2\\ \frac{\mathrm{λ}}{4}+x=\frac{d^2}{\mathrm{λ}}\\ x=\frac{d^2}{\mathrm{λ}}−\frac{\mathrm{λ}}{4}\end{array}$

In general, if $\text{Δ}L=n\mathrm{λ}$where n is some multiplier,

$\text{Δ}L=\sqrt{d^2+x^2}−x$

Substitute $n\mathrm{λ}$for $\text{Δ}L$into the above equation,

$\begin{array}{c}n\mathrm{λ}=\sqrt{d^2+x^2}−x\\ x^2+{\left(n\mathrm{λ}\right)}^2+xn\mathrm{λ}=d^2+x^2\\ {\left(n\mathrm{λ}\right)}^2+2xn\mathrm{λ}=d^2\\ n\mathrm{λ}+2x=d^2/n\mathrm{λ}\end{array}$

Express the above equation in terms of x

$\begin{array}{c}n\mathrm{λ}+2x=d^2/n\mathrm{λ}\\ x=\frac{d^2}{2n\mathrm{λ}}−\frac{1}{2}n\mathrm{λ}\\ x=\frac{d^2}{2n\mathrm{λ}}−\frac{1}{2}n\mathrm{λ}\\ x=\frac{{16}^2}{2n×2}−\frac{1}{2}n×2\\ x=\frac{256}{4n}−n\end{array}$

$⇒x=\frac{64}{n}−n$

Now, phase difference $\text{Δ}L=0.50\mathrm{λ},$i.e.$n=0.50$, Find distance x as,

Substitute $0.50$for into the above equation,

$\begin{array}{l}x=\frac{64}{n}−n\\ =\frac{64}{0.5}−0.5\\ =127.\text{5m}\\ =1\text{28m}\end{array}$

Hence, at distance $x=128\text{m,}$the waves have a phase difference of$0.50\mathrm{λ}$ .

Now, phase difference$\text{Δ}L=1.00\mathrm{λ},$i.e.$n=\mathrm{1.00.}$,Find distance x as,

$\begin{array}{c}x=\frac{64}{n}−n\\ =\frac{64}{1.00}−1\\ =63\text{m}\end{array}$

Hence, at distance $x=63.\text{0m,}$the waves have a phase difference of $1.00\mathrm{λ}$.

Now, phase difference$\text{Δ}L=1.50\mathrm{λ},$i.e.$n=\mathrm{1.50.}$,Find distance x as,

$\begin{array}{l}x=\frac{64}{n}−n\\ =\frac{64}{1.50}−1.50\\ =41.2\text{m}\end{array}$

Hence, at distance$x=41.\text{2m,}$ the waves have a phase difference of$1.50\mathrm{λ}$