Two sound waves having a phase difference of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The relationship between path difference $(\Delta x)$ and phase difference $(\phi)$ is given by the formula: $\Delta x = \frac{\lambda}{2\pi} 	im

Vedclass · Accepted Answer

$\lambda / 6$

Vedclass · Answer

(C) The relationship between path difference $(\Delta x)$ and phase difference $(\phi)$ is given by the formula: $\Delta x = \frac{\lambda}{2\pi} 	imes \phi$. Given phase difference $\phi = 60^{\circ}$. Convert the phase difference into radians: $\phi = 60^{\circ} 	imes \frac{\pi}{180^{\circ}} = \frac{\pi}{3} 	ext{ radians}$. Substitute the value of $\phi$ into the formula: $\Delta x = \frac{\lambda}{2\pi} 	imes \frac{\pi}{3}$. Simplifying the expression: $\Delta x = \frac{\lambda}{6}$. Therefore,the path difference is $\lambda / 6$.

Two sound waves having a phase difference of $60^{\circ}$ have a path difference of

Similar Questions

Write the definition,unit,and dimensional formula of amplitude.

For a wave,the path difference is $x$. The corresponding phase difference is ...

$A$ plane wave $y=a \sin (\omega t-k x)$ propagates through a stretched string. The particle velocity versus $x$ graph at $t=0$ is

The displacement of a wave is given by $y=0.002 \sin (100t + x)$,where $x$ and $y$ are in meters and $t$ is in seconds. This represents a wave:

The path difference between two waves,represented by $y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}\right)$ and $y_2 = a_2 \cos \left(\omega t - \frac{2 \pi x}{\lambda} + \phi\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module