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(B) Given equation: $y=0.05 \sin \pi(2 t-0.02 x)$ Expanding the equation: $y=0.05 \sin (2 \pi t - 0.02 \pi x)$ Comparing this with the standard wave e

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$100 \ m, 100 \ ms^{-1}$

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(B) Given equation: $y=0.05 \sin \pi(2 t-0.02 x)$ Expanding the equation: $y=0.05 \sin (2 \pi t - 0.02 \pi x)$ Comparing this with the standard wave equation $y=a \sin (\omega t - k x)$: Angular frequency $\omega = 2 \pi \ rad/s$ Wave number $k = 0.02 \pi \ m^{-1}$ The minimum distance between two particles in phase is the wavelength $\lambda$. Since $k = \frac{2 \pi}{\lambda}$,we have $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{0.02 \pi} = \frac{2}{0.02} = 100 \ m$. The wave velocity $v$ is given by $v = \frac{\omega}{k}$. $v = \frac{2 \pi}{0.02 \pi} = \frac{2}{0.02} = 100 \ m/s$.

The equation of a transverse wave is given by $y=0.05 \sin \pi(2 t-0.02 x)$,where $x, y$ are in metre and $t$ is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively

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