Question

The path difference between two waves

y₁ = a₁ sin $\left( \omega t - \frac{2\pi x}{\lambda} \right)$and y₂ = a₂cos$\left( \omega t - \frac{2\pi x}{\lambda} + \varphi \right)$is –

• A. $\frac{\lambda}{2\pi}$ (f)
• B. $\frac { \lambda } { 2 \pi }$ $\left( \varphi + \frac{\pi}{2} \right)$
• C. $\frac { 2 \pi } { \lambda }$ $\left( \varphi - \frac{\pi}{2} \right)$
• D. $\frac{2\pi}{\lambda}$ (f)

Answer 1

Answer: (B)

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

Answer 2

Q phase difference in y₁ & y₂ is = $\left( \frac{\pi}{2} + \varphi \right)$

Dx = $\frac { \lambda } { 2 \pi }$ $\left\lbrack \frac{\pi}{2} + \varphi \right\rbrack$