Question 381: Waves

Question

The path difference between two waves, represented by $$\mathrm{y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$$ and $$y_2=a_2 \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)$$ is

Accepted Answer

$$\begin{aligned}
& \mathrm{y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}ight) \
& \mathrm{y}_2=\mathrm{a}_2 \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phiight)
\end{aligned}$$
$$\mathrm{y}_2$$ can also be written as
$$\begin{aligned}
\mathrm{y}_2 & =\mathrm{a}_2 \sin \left[\frac{\pi}{2}+\left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phiight)ight] \
& =\mathrm{a}_2 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi+\frac{\pi}{2}ight)
\end{aligned}$$
The phase difference between the two waves is
Path difference $$=\frac{\lambda}{2 \pi} 	imes$$ Phase difference
Path difference $$=\frac{\lambda}{2 \pi} 	imes\left(\phi+\frac{\pi}{2}ight)$$

Answer

$$\frac{\lambda}{2 \pi}(\phi)$$

Answer

$$\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$$

Answer

$$\frac{2 \pi}{\lambda}\left(\phi-\frac{\pi}{2}\right)$$

Answer

$$\frac{2 \pi}{\lambda}(\phi)$$

Waves

The path difference between two waves, represented by $\mathrm{y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right) and y_2=a_2 \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)$ is

Choose the correct answer: