The equation of a simple harmonic progressive | vedclass

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

Question

(A) The given equation of the wave is $y=A \sin (100 \pi t-3 x)$.Comparing this with the standard wave equation $y=A \sin (\omega t-kx)$,we get the wa

Vedclass · Accepted Answer

$\frac{\pi}{9} \ m$

Vedclass · Answer

(A) The given equation of the wave is $y=A \sin (100 \pi t-3 x)$.Comparing this with the standard wave equation $y=A \sin (\omega t-kx)$,we get the wave number $k=3 \ m^{-1}$.The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by $\Delta \phi = k \cdot \Delta x$.Given the phase difference $\Delta \phi = \frac{\pi}{3}$.Substituting the values,we have $\frac{\pi}{3} = 3 \cdot \Delta x$.Solving for the distance $\Delta x$,we get $\Delta x = \frac{\pi}{3 	imes 3} = \frac{\pi}{9} \ m$.

The equation of a simple harmonic progressive wave is given by $y=A \sin (100 \pi t-3 x)$. Find the distance between $2$ particles having a phase difference of $\frac{\pi}{3}$.

Similar Questions

$A$ wave is given by $Y = 3 \sin 2 \pi \left( \frac{t}{0.04} - \frac{x}{0.01} \right)$ where $Y$ is in $cm$. The frequency of the wave and the maximum acceleration will be $(\pi^2 = 10)$.

$A$ transverse wave is represented by the equation $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$. For what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity?

$A$ plane progressive wave is given by $y = 2 \cos 2 \pi (330 t - x) \ m$. The frequency of the wave is: (in $Hz$)

$A$ 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:

$A$ transverse wave is given by $y = A \sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to $4$ times the wave velocity when:

Vedclass Test Series

Exam Paper Generator

Online Exam Module