A travelling acoustic wave of frequency 500 H | vedclass

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

Question

(C) The frequency of the wave is $f = 500 Hz$ and the velocity is $v = 300 ms^{-1}$.First, we calculate the wavelength $\lambda$ using the relation $v

Vedclass · Accepted Answer

$10 cm$

Vedclass · Answer

(C) The frequency of the wave is $f = 500 Hz$ and the velocity is $v = 300 ms^{-1}$.First, we calculate the wavelength $\lambda$ using the relation $v = f \lambda$:$\lambda = \frac{v}{f} = \frac{300}{500} = 0.6 m = 60 cm$.The phase difference $\Delta \phi$ is given as $60^{\circ}$, which is $\frac{\pi}{3}$ radians.The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Substituting the values: $\frac{\pi}{3} = \frac{2 \pi}{60 cm} \Delta x$.Solving for $\Delta x$: $\Delta x = \frac{\pi}{3} 	imes \frac{60 cm}{2 \pi} = \frac{60}{6} cm = 10 cm$.

$A$ travelling acoustic wave of frequency $500 Hz$ is moving along the positive $x$-direction with a velocity of $300 ms^{-1}$. The phase difference between two points $x_{1}$ and $x_{2}$ is $60^{\circ}$. The minimum separation between the two points is:

Similar Questions

The path difference between two waves $Y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}\right)$ and $Y_2 = a_2 \cos \left(\omega t - \frac{2 \pi x}{\lambda} + \phi\right)$ is

$A$ wave travelling along the $x$-axis is described by the equation $y(x, t) = 0.005 \cos(\alpha x - \beta t)$. If the wavelength and the time period of the wave are $0.08 \ m$ and $2.0 \ s$ respectively,then $\alpha$ and $\beta$ in appropriate units are:

$A$ plane wave is represented by $x = 1.2 \sin(314t + 12.56y)$,where $x$ and $y$ are distances measured in meters and $t$ is time in seconds. This wave has:

The figure shows the shape of a part of a long string in which transverse waves are produced by attaching one end of the string to a tuning fork of frequency $250 \ Hz$. What is the velocity of the waves in $ms^{-1}$?

Two points are located at a distance of $10\; m$ and $15 \;m$ from the source of oscillation. The period of oscillation is $0.05 \;s$ and the velocity of the wave is $300 \;m/s$. What is the phase difference between the oscillations of the two points?

Vedclass Test Series

Exam Paper Generator

Online Exam Module