Two points are located at a distance of 10; m | vedclass

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(D) Given:Distance of first point,$x_1 = 10\;m$Distance of second point,$x_2 = 15\;m$Period of oscillation,$T = 0.05\;s$Velocity of wave,$v = 300\;m/s

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(D) Given:Distance of first point,$x_1 = 10\;m$Distance of second point,$x_2 = 15\;m$Period of oscillation,$T = 0.05\;s$Velocity of wave,$v = 300\;m/s$Path difference between the two points is:$\Delta x = x_2 - x_1 = 15 - 10 = 5\;m$First,calculate the wavelength $\lambda$ using the formula $v = f \lambda$,where $f = \frac{1}{T}$:$f = \frac{1}{0.05} = 20\;Hz$$\lambda = \frac{v}{f} = \frac{300}{20} = 15\;m$The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula:$\Delta \phi = \frac{2\pi}{\lambda} 	imes \Delta x$Substituting the values:$\Delta \phi = \frac{2\pi}{15} 	imes 5 = \frac{2\pi}{3}$

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?

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